Pra-U STPM Maths(T) Semester 3 2022 CC039332c

46 Mathematics Semester 3 STPM Chapter 1 Data Description 1 Exercise 1.7 1. The following table shows the frequency distribution of families with a certain number of members. Number of members 1 2 3 4 5 6 7 8 Number of families 10 59 102 98 121 66 25 4 Calculate the standard deviation for this data. 2. The following histogram shows the number of patients treated at a clinic everyday during a certain period of time. 50 40 30 20 10 0 39.5 49.5 59.5 69.5 79.5 89.5 99.5 40 18 27 48 36 15 Days Number of patients Calculate standard deviation of the number of patients treated at the clinic. 3. The engine capacities of 112 cars are summarised in the table below. Find the standard deviation. Engine capacities (x cm3 ) Number of cars 1 800  x  1000 5 1000  x  1200 17 1200  x  1400 26 1400  x  1600 38 1600  x  1800 13 1800  x  2000 8 2000  x  2200 5 4. The frequency table below shows the distances of a group of workers’ houses from their place of work. Distance (x km) Frequency 10  x  21 32 12  x  41 91 14  x  61 41 16  x  81 28 18  x  10 13 10  x  12 10 Calculate the standard deviation.

47 Mathematics Semester 3 STPM Chapter 1 Data Description 1 5. The cumulative frequency table shows the numbers of years a group of workers continue to live after retirement. Number of years Number of people  01 0  21 12  41 36  61 88  81 122  10 152  12 176  14 191  16 205 Besides that, there are 3 workers who live for 18 years and 2 workers who live for 19 years after retiring. Find the standard deviation of the numbers of years the workers live on after retirement. 6. The table below shows the numbers of customers served and their waiting periods at two counters in a post office. The numbers of customers served are given in cumulative form. Waiting period (minutes) Number of customers served at counter X Number of customers served at counter Y  01 0 0  51 8 6  10 18 11  15 42 27  20 61 49  25 78 67  30 90 78  35 90 83  40 90 90 Find the standard deviation of the waiting periods for each of the counters. 7. A factory supplies chilli sauce to provision shops. The numbers of bottles of chilli sauce supplied during a period of one month are given in the table below. Number of bottles of chilli sauce 50 100 150 200 250 300 Number of shops 58 65 91 52 29 5 Find the mean and standard deviation.

48 Mathematics Semester 3 STPM Chapter 1 Data Description 1 8. The table below shows the numbers of companies with different investment capitals. Investment capital (RMy) Number of companies 200 000  y  300 00000 5 300 000  y  400 00000 16 400 000  y  500 00000 21 500 000  y  600 00000 23 600 000  y  700 00000 18 700 000  y  800 00000 9 800 000  y  900 00000 4 900 000  y  1 000 000 2 Use the coding method to find the standard deviation for the investment capitals of the companies. 9. A survey is carried out to find out the amounts of money borrowed by 200 people from a bank to buy a house. The results are summarised in the following table. Amount borrowed (RM) Number of borrowers 20 000 – 39 9991 5 40 000 – 59 9991 33 60 000 – 79 9991 57 80 000 – 99 9991 32 100 000 – 119 999 36 120 000 – 139 999 27 140 000 – 159 999 10 Use the coding method to calculate the mean and standard deviation of the amounts of money borrowed to buy a house. 10. For a particular set of data, n = 100, ∑(x – 40) = 112.4, ∑(x – 40)2 = 218.6. Find the mean and standard deviation of x. 11. Given that ∑f = 50, ∑f(x – 80) = 140, ∑f(x – 80)2 = 2615, find the mean and standard deviation of x.

49 Mathematics Semester 3 STPM Chapter 1 Data Description 1 1.4 The Shape of a Distribution If a distribution is represented by a histogram or a frequency curve, we can see the general shape of its distribution. (a) Symmetrical distribution (bell-shaped) Frequency Variable Mode = Median = Mean Frequency Variable Figure 1.4 In a symmetrical distribution (Figure 1.4), the mode, median and mean are all of the same value. This distribution is also known as the normal distribution. We will study the normal distribution in more detail in Chapter 3. (b) Positively skewed distribution (skewed to the right) Frequency Variable Mode Median Mean Frequency Variable Figure 1.5 In a positively skewed distribution (Figure 1.5), mode  median  mean. This type of distribution occurs in our everyday life. For example, when we study the number of children per family or the wages of workers in an electronics factory. (c) Negatively skewed distribution (skewed to the left) Frequency Variable Mode Median Mean Frequency Variable Figure 1.6

50 Mathematics Semester 3 STPM Chapter 1 Data Description 1 In a negatively skewed distribution (Figure 1.6), mean  median  mode. This distribution can occur when we study the lowest daily temperatures in a certain month or the reaction time of a chemical reaction. Note: For distributions which are largely skewed, whether positively or negatively, median is a more appropriate measure of central tendency. For symmetrical normal distributions or almost symmetrical distributions, mean is an appropriate measure of central tendency. Exercise 1.8 1. A farmer records the number of eggs collected everyday from a farm for 65 days. The results are summarised in the table below. Number of eggs 10 11 12 13 14 15 16 17 Frequency 2 3 6 9 10 15 12 8 Determine the mode, median and mean of the above distribution. Draw a line graph to represent the data and comment on the shape of the graph. 2. The table below shows the times (to the nearest minute) used to read fiction books in a day by a group of secondary students. Time (minutes) Number of students 20 – 29 10 30 – 34 18 35 – 39 26 40 – 49 22 50 – 59 14 60 – 74 9 75 – 95 4 (a) Represent this data on a histogram. (b) Comment on the distribution. 3. The frequency distribution of the lengths of a type of leaf (in mm) measured by 40 pupils is given below. Length (mm) Frequency 30 – 39 4 40 – 49 8 50 – 59 10 60 – 69 9 70 – 79 5 80 – 89 3 90 – 99 1

51 Mathematics Semester 3 STPM Chapter 1 Data Description 1 (a) Draw a histogram to illustrate the above data. (b) Determine whether the mean  mode or mode  mean. (c) Give a comment on the data distribution. 4. The table below shows the results of a survey to determine the lifespans of 200 bulbs taken at random from a factory. Lifespan (hours) Number of bulbs 640 – 659 8 660 – 669 13 670 – 679 15 680 – 689 39 690 – 694 30 695 – 699 43 700 – 704 37 705 – 714 10 715 – 734 4 (a) Draw a histogram to represent the data. (b) State whether this distribution is positively or negatively skewed. (c) Determine whether mean or median is more suitable for representing the data. 5. The blood pressures for 52 adults are summarised in the following table. Blood pressure (mm Hg) 95 – 104 105 – 109 110 – 114 115 – 119 120 – 124 125 – 129 130 – 134 135 – 139 140 – 170 Frequency 1 4 5 8 12 7 6 3 6 (a) Calculate the mean and median. (b) State whether this distribution is positively or negatively skewed. Pearson coefficient of skewness The degree of skewness of a distribution is measured by calculating how far the distribution is pulled in one direction. The skewness of a distribution is either positive or negative. Pearson coefficient of skewness is determined by using the following formula. Pearson coefficient of skewness = mean – mode standard deviation If mean = mode, the Pearson coefficient of skewness is zero and the distribution is symmetrical. If mean . mode, the Pearson coefficient of skewness is positive and the distribution is positively skewed. If mean , mode, the Pearson coefficient of skewness is negative and the distribution is negatively skewed. Using the approximation, mean – mode = 3(mean – median), Pearson coefficient of skewness = 3(mean – median) standard deviation . Pearson Coefficient of Skewness INFO

52 Mathematics Semester 3 STPM Chapter 1 Data Description 1 Figure 1.7 shows examples of distributions with positive and negative Pearson coefficients of skewness. Frequency Variable Frequency Variable Pearson coefficient of skewness = 2.0 Pearson coefficient of skewness = –1.9 Figure 1.7 Example 31 Calculate the Pearson coefficient of skewness for each of the following cases. (a) A frequency distribution as follows: x 20 30 40 50 60 70 80 f 11 13 18 28 14 5 3 (b) A distribution with mean = 18, median = 22 and standard deviation = 5. Solution: (a) Mode = 50, mean, x = ∑fx ∑f = 11(20) + 13(30) + 18(40) + 28(50) + 14(60) + 5(70) + 3(80) 11 + 13 + 18 + 28 + 14 + 5 + 3 = 4 160 92 = 45.22 Variance, s² = ∑fx 2 ∑f – x 2 = 11(20)² + 13(30)² + 18(40)² + 28(50)² + 14(60)² + 5(70)² + 3(80)² 92 – 45.222 = 226.89 Standard deviation, s = 15.06 Pearson coefficient of skewness = mean – mode standard deviation = 45.22 – 50 15.06 = –0.317 (b) Pearson coefficient of skewness = 3(mean – median) standard deviation = 3(18 – 22) 5 = –2.4

53 Mathematics Semester 3 STPM Chapter 1 Data Description 1 Exercise 1.9 1. For a distribution, the mean is 78, the mode is 67 and the variance is 16. Calculate the Pearson coefficient of skewness. Interpret your answer. 2. Marks in an examination are represented by the following stem-and-leaf diagram. Stem Leaf 1 2 3 4 5 6 7 8 9 5 3 8 9 4 7 1 2 7 8 2 3 5 5 5 7 8 4 6 6 1 8 2 Key: 6 | 2 means 62 marks Find the Pearson coefficient of skewness. 3. Calculate the Pearson coefficient of skewness for each of the following frequency distributions. (a) f 2 4 7 6 3 2 1 1 0 x 12 13 14 15 16 17 18 19 20 (b) f 1 2 3 5 7 9 5 2 x 24 25 26 27 28 29 30 31 Box-and-whisker plots (boxplots) Box-and-whisker plots show the spread of a distribution by using the smallest value, largest value, first quartile, third quartile and median. Boxplots can be displayed horizontally (Figure 1.8) or vertically (Figure 1.9). 0 10 20 30 40 50 60 Largest value Smallest value Third quartile, Q3 First quartile, Q1 Median, Q2 Figure 1.8 Box-andwhisker Plots INFO

54 Mathematics Semester 3 STPM Chapter 1 Data Description 1 60 50 40 30 20 10 0 Smallest value Largest value First quartile, Q1 Median, Q2 Third quartile, Q3 Figure 1.9 The ‘box’ starts from Q1 up to Q3 and contains 50% of the data in the middle of the distribution. The ‘whisker’ starts from the box to the smallest value and also from the box to the largest value. The ‘whisker’ displays the range of the data. The boxplot for a symmetrical distribution is shown in Figure 1.10. The median is in the centre of the box and the ‘whiskers’ are of the same length. Notice that Q3 – Q2 = Q2 – Q1 Q1 Q2 Q3 Figure 1.10 The boxplot for a positively skewed distribution is shown in Figure 1.11. The median is nearer to the first quartile and the left ‘whisker’ is shorter than the right ‘whisker’. Notice that Q3 – Q2 . Q2 – Q1 Q1 Q2 Q3 Figure 1.11

55 Mathematics Semester 3 STPM Chapter 1 Data Description 1 The boxplot for a negatively skewed distribution is shown in Figure 1.12. The median is nearer to the third quartile and the left ‘whisker’ is longer than the right ‘whisker’. Notice that Q3 – Q2 , Q2 – Q1 Q1 Q2 Q3 Figure 1.12 Example 32 The marks of Mathematics and Biology for students in a class are summarised below. Subject Minimum Maximum Median First quartile Third quartile Mathematics 10 90 60 45 70 Biology 35 85 60 48 72 Draw two boxplots for this data and give comments regarding the distribution of marks for Mathematics and Biology. Solution: 0 10 Mathematics Biology 20 30 40 50 60 70 80 90 100 Comment: The median marks of both subjects are the same but the Mathematics marks show a bigger range and are skewed to the left. The Biology marks have a smaller range compared to Mathematics marks and are symmetrical. Example 33 The frequency distribution of the ages (in years) of 48 used cars for sale is given below. Age 1 2 3 4 5 6 7 8 9 Frequency 7 12 8 6 5 4 3 2 1 (a) Find all the quartiles. (b) Draw a boxplot to represent the data. (c) State the shape of the distribution based on your boxplot. Give a reason.

56 Mathematics Semester 3 STPM Chapter 1 Data Description 1 Solution: (a) (i) Median = average of the 24th and 25th observations = 1 2 (3 + 3) years = 3 years (ii) First quartile = average of the 12th and 13th observations = 2 years Third quartile = average of the 36th and 37th observations = 5 years (b) 1 2 3 4 5 6 7 8 9 Age (years) (c) The distribution of the cars’ ages is skewed to the right, since Q3 – Q2 . Q2 – Q1 . Box-and-whisker plots with outliers Sometimes, values which are unusually small or large occur in a data set. The unusual values occur probably because of an error in recording the data. As a guide, values which are 1.5 times the interquartile range more than the third quartile or less than the first quartile are called ‘outliers’. Q1 Q3 Smallest value inside boundary Largest value inside boundary Lower boundary Upper boundary Outlier Outlier 1.5 (Q3 – Q1) 1.5 (Q3 – Q1) Figure 1.13 Example 34 The following stemplot shows the maximum temperature of each month in a town for 23 months. Identify any ‘outliers’ and draw a boxplot. Stem Leaf 1 2 3 4 5 6 7 6 7 0 2 2 3 5 7 8 8 8 9 9 2 3 3 4 4 4 4 9 5 8 Key: 5 | 9 means 59°C

57 Mathematics Semester 3 STPM Chapter 1 Data Description 1 Solution: First quartile, Q1 = 23°C Median, Q2 = 39°C Third quartile, Q3 = 44°C Lower boundary = Q1 – 1.5(Q3 – Q1 ) = 23 – 1.5(44 – 23) = 75.5°C Upper boundary = Q3 + 1.5(Q3 – Q1 ) = 44 + 1.5(44 – 23) = –8.5°C Hence, the temperature that may be recorded wrongly is 78°C, considered an outlier. 16 23 39 44 65 78 X Exercise 1.10 1. Draw a boxplot to represent each of the following frequency distributions. (a) x 1 2 3 4 5 f 6 10 6 2 1 (b) x 10 11 12 13 14 15 16 f 2 6 12 14 13 5 3 (c) x 0 1 2 3 4 5 6 7 f 2 5 6 7 9 14 12 8 (d) x 46 47 48 49 50 51 52 f 5 21 35 20 10 5 4 2. The times a Pos Malaysia van collects mails from a public letter box for 27 days are recorded below. 8.10 8.12 8.13 8.19 8.10 8.25 8.17 8.48 8.02 8.45 8.53 8.02 8.05 8.31 8.47 8.29 8.09 8.29 8.29 8.15 8.19 8.30 8.22 8.01 8.02 8.21 8.22 (a) Display the data in a stem-and-leaf diagram. (b) Draw box-and-whisker plot to represent the data.

58 Mathematics Semester 3 STPM Chapter 1 Data Description 1 3. The table below shows the duration of each telephone call (in minutes) in an office for 50 calls. Duration of a call (minutes) Number of calls  11 7 1 – 21 12 2 – 31 18 3 – 51 7 5 – 10 5  10 1 (a) Estimate the median, first quartile and third quartile. (b) Draw a boxplot. (c) Comment on the shape of the distribution. 4. The times taken (in minutes) by 35 people to prepare a special meal each are recorded below. 12 25 79 54 62 82 73 37 71 49 67 76 68 54 50 87 43 40 80 43 88 66 73 48 145 24 37 121 61 74 81 55 43 49 134 Identify any outliers and draw a boxplot. 5. The following boxplots show the reaction times for two groups of students A and B. 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 Group A students Group B students Comment on the distributions. 6. The cumulative frequency curves of chemistry marks for 100 girls and 100 boys are shown below. 0 10 10 20 30 40 50 60 70 80 20 30 40 50 60 70 80 90 100 Cumulative frequency Marks Girls Boys Draw boxplots to represent the chemistry marks for the girls and the boys. Comment on the distributions.

59 Mathematics Semester 3 STPM Chapter 1 Data Description 1 7. The back-to-back stemplot below shows the numbers of lots of shares of a company sold by a share broker in the months of January and July. January July 3 2 4 2 1 6 5 3 0 8 8 6 2 7 6 4 4 3 3 3 1 5 4 2 2 7 1 0 9 2 1 0 1 2 3 4 5 6 7 8 9 10 2 2 3 4 4 4 4 6 7 8 0 4 5 5 6 6 6 1 1 2 3 7 3 5 5 6 8 3 9 1 0 Key: 1 | 6 means 61 lots Key: 4 | 5 means 45 lots (a) Find the median, first quartile and third quartile for each month and construct boxplots for the two months. (b) Comment on the distributions of the numbers of lots of shares sold in the months of January and July. Summary 1. (a) Discrete data can only take exact values. For example, the number of workers in a restaurant. (b) Continuous data cannot take exact values but lie in an interval. For example, the height of a student. 2. For an ungrouped data set x1 , x2 , …, xn arranged in ascending order, (a) mode is the observation that occurs the most often. (b) if n is an odd integer, median = 1 n + 1 2 2th observation. if n is an even integer, median = average of the 1 n 2 2th and 1 n 2 + 12th observations. (c) mean, x = x1 + x2 + … + xn n = ∑x n (d) range = largest observation – smallest observation. (e) interquartile range = Q3 – Q1 (f) standard deviation = ∑(x – x ) 2 n = ∑x2 n – 1 ∑x n 2 2

60 Mathematics Semester 3 STPM Chapter 1 Data Description 1 3. For a frequency distribution where the mid-points of the classes are x1 , x2 , x3 , …, xn with respective frequencies f 1 , f 2 , f 3 , …, f n, (a) median, m = Lm + 3 1 2 (∑f ) – Fm – 1 f m 4 c where Lm = lower boundary of the median class, Fm – 1 = cumulative frequency before the median class, f m = frequency of the median class, c = width of the median class. (b) Q1 Q1 = First quartile Q2 = Median Q3 = Third quartile Q2 Q3 Cumulative frequency Variable 1 (�f) _ 4 1 (�f) _ 2 3 (�f) _ 4 (c) mean, x = ∑fx ∑f (d) interquartile range = Q3 – Q1 (e) standard deviation = ∑(x – x ) 2 n = ∑fx2 ∑f – 1 ∑fx ∑f 2 2 4. The coding formula is y = x – k h where k = assumed mean and h = scaling factor. (a) x = hy + k, where x = mean of x, y = mean of y (b) s x 2 = h2 s y 2 , where s x 2 = variance of x-values, s y 2 = variance of y-values 5. Pearson coefficient of skewness = mean – mode standard deviation or 3(mean – median) standard deviation . 6. Quartile coefficient of skewness (a) If Q3 – Q2 = Q2 – Q1 , then the quartile skewness = 0 and the distribution is normal. (b) If Q3 – Q2 . Q2 – Q1 , then the quartile skewness . 0 and the distribution is positively skewed. (c) If Q3 – Q2 , Q2 – Q1 , then the quartile skewness , 0 and the distribution is negatively skewed.

61 Mathematics Semester 3 STPM Chapter 1 Data Description 1 STPM PRACTICE 1 1. The masses (to the nearest gram) 12 bars of soap are as follows. 174 164 182 169 171 187 176 177 168 171 180 175 (a) State the mode and range. (b) Find the median and interquartile range. 2. The moisture content of paddy in each of 30 sacks is measured. The results (percentage of water in the paddy) are as follows: 15.0 16.3 13.1 12.9 14.2 15.7 15.4 15.0 15.1 15.7 14.2 13.4 14.7 14.9 15.3 12.8 15.9 14.6 14.4 14.2 14.7 14.5 15.1 16.1 15.4 14.9 15.0 14.0 14.3 14.4 (a) Display the data in a stem-and-leaf diagram. (b) State the mode(s) and range. (c) Determine the median and interquartile range. 3. The ages (to the nearest year) of 36 customers who register their names to book Proton Perdana in a certain month are as follows: 28 35 31 32 31 25 34 26 37 33 30 29 30 36 48 44 32 37 41 37 39 60 51 41 37 37 31 28 43 32 35 41 30 37 50 43 (a) Copy and complete the stemplot below. 25 30 35 3 1 2 0 (b) State the mode and range. (c) Determine the median and interquartile range. 4. Each member in a sample consisting of 100 students is required to record the total time spent watching television in a period of 3 weeks. The frequency table below is obtained. Total time (hours) 5  x 15 15  x  25 25  x  35 35  x  45 45  x  55 55  x  65 Number of members 10 18 30 20 16 6 Calculate the median and interquartile range.

62 Mathematics Semester 3 STPM Chapter 1 Data Description 1 5. The table below shows the lifespans of 200 electric bulbs taken from a factory. Lifespan (days) Frequency 690 – 709 3 710 – 719 7 720 – 729 15 730 – 739 38 740 – 744 41 745 – 749 35 750 – 754 21 755 – 759 16 760 – 769 14 770 – 789 10 Calculate estimates of the median and the interquartile range of the lifespans of the electric bulbs. 6. The cumulative frequency table below shows the age distribution of 184 people who board a bus between 0800 hours and 0900 hours one Monday morning. Age (years)  5  10  15  20  25 Cumulative frequency 0 58 123 152 167 Age (years)  30  35  40  45  50 Cumulative frequency 176 180 182 183 184 (a) Plot a cumulative frequency curve. (b) Estimate the median and interquartile range. 7. The following table shows the heights (in cm) of 400 students. Height (cm) Cumulative frequency  100 0  110 27  120 85  130 215  140 320  150 370  160 395  170 400 Plot a cumulative frequency curve. Use your curve to estimate the median and interquartile range.

63 Mathematics Semester 3 STPM Chapter 1 Data Description 1 8. Everyday, a train leaves town A for town B at 0828 hours. The time taken for this journey is recorded for a certain period. The results are shown in the following table. Time (minutes) Frequency –80 0 –85 6 –90 12 –95 22 –100 31 –105 15 –110 7 –115 4 –120 2 –125 1  125 0 (The interval –90 represents times which are more than 85 minutes but not more than 90 minutes.) (a) Plot a cumulative frequency curve. (b) Determine the median and interquartile range. (c) Find the number of times a train arrives at town B between 1000 hours and 1015 hours. 9. The following table shows the masses of garbage in a week for 156 households. Mass (kg) Number of households 3.00 – 3.19 22 3.20 – 3.39 34 3.40 – 3.59 42 3.60 – 3.79 16 3.80 – 3.99 12 4.00 – 4.19 9 4.20 – 4.39 10 4.40 – 4.59 4 4.60 – 4.79 4 4.80 – 4.99 3 Construct a cumulative frequency table and draw a cumulative frequency curve. Hence, estimate (a) the median, (b) the percentage of households which have between 3.46 kg and 3.85 kg of garbage in a week. 10. The mass (in grams) of each of 10 tins of red beans is measured. The results are recorded as follows: 512, 515, 499, 528, 519, 510, 507, 522, 530, 514 (a) Calculate the mean and standard deviation. (b) Find the percentage of observations which are more than the mean.

64 Mathematics Semester 3 STPM Chapter 1 Data Description 1 11. Two machines, A and B, are used for packing 200 g of biscuits. A sample which contains 10 packets is taken from each machine and the mass (to the nearest gram) of each packet is measured. The results are as follows: Machine A: 196, 198, 198, 199, 200, 200, 201, 201, 202, 205 Machine B: 192, 194, 195, 198, 200, 201, 203, 204, 206, 207 Find the mean and standard deviation of the masses of the packets for each machine. Comment on your answers. 12. The pH values of 10 samples of soil taken from a piece of land are shown below. 6.0 4.5 5.7 5.8 6.1 7.3 4.8 5.6 5.8 4.3 Find the percentage of the samples of soil having pH values between the mean and the median. 13. A regular customer of a hardware shop records the numbers of customers in the shop when he is there as follows: Number of customers Frequency 0 15 1 34 2 27 3 14 4 10 He also records the average waiting times before he served as follows: Number of customers Average waiting time (minutes) 0 0 1 2 2 6 3 9 4 12 (a) Find the mean number of customers. (b) Find the mean waiting time. 14. The lifespans (in hours) for a certain type of batteries are measured for a sample of 100 batteries. The results are given below. Lifespan (hours) 90 95 100 105 110 115 120 125 130 Frequency 2 17 30 21 15 9 3 2 1 (a) Calculate the median lifespan of the batteries. (b) Calculate the mean lifespan of the batteries and the standard deviation. (c) Find the fraction of batteries that have lifespans within one standard deviation from the mean.

65 Mathematics Semester 3 STPM Chapter 1 Data Description 1 15. The ages and numbers of workers in the agricultural sector in a certain year are shown in the table below. Age (years) Number of workers (in thousands) 15 – 19 66 20 – 24 65 25 – 29 56 30 – 34 50 35 – 39 42 40 – 44 37 45 – 49 35 50 – 54 30 55 – 59 24 60 – 64 22 (a) Calculate the mean and standard deviation of the ages of the workers. (b) Find the percentage of workers whose ages are within one standard deviation from the mean. 16. A machine is designed to cut metal rods of length 120 cm. A sample of 100 metal rods is taken and the length of each rod is measured. The results are recorded below. Length of metal rod (x cm) Number of rods 119.996  x  119.998 10 119.998  x  120.000 50 120.000  x  120.002 30 120.002  x  120.004 10 (a) Display the above results in a cumulative frequency graph. (b) Estimate the median for this grouped frequency distribution. (c) Determine the mean for this grouped frequency distribution. Give your answer correct to four decimal places. (d) Calculate estimates of the mean and the standard deviation. Give your answers correct to three decimal places. 17. The mean weight of 11 students from a college is 58 kg. The weights (in kg) of 10 of the students are as follows. 55 60 53 62 64 57 56 59 65 58 Calculate the standard deviation of the weights of the 11 students.

66 Mathematics Semester 3 STPM Chapter 1 Data Description 1 18. An automatic machine is set to produce steel ball bearings with diameter 15.85 mm. Over a long interval of time, the percentage of ball bearings within certain diameter ranges are shown in the table below. Diameter (x mm) Percentage (%) x  15.55 1 15.55  x  15.65 3 15.65  x  15.75 19 15.75  x  15.85 43 15.85  x  15.95 25 15.95  x  16.05 7 16.05 < x 2 Total 100 Calculate the mean diameter of the steel ball bearings produced by the machine. A sample of 12 steel ball bearings is taken after the machine is serviced. The diameters of these ball bearings (in mm) are recorded as follows. 15.79 15.84 16.12 16.03 15.71 15.93 15.81 15.73 15.80 15.84 15.92 15.97 Calculate the mean for this sample and comment on the performance of the machine after servicing. 19. 80 batteries produced in the morning shift and 100 batteries produced in the afternoon shift in a factory had their lifespans tested. The following table gives the means and standard deviations of the lifespans of the two groups of batteries. Battery Mean (hours) Standard deviation (hours) Morning shift 16.7 2.3 Afternoon shift 18.2 1.9 Calculate the mean and standard deviation of the lifespans of all the batteries. 20. Show that (from basic definition) the variance of a set of observations y1 , y2 , y3 , …, yn with mean y can be determined by using 1 n n ∑ i=1 yi 2 – y 2 . The heights of 20 football players are recorded. The mean height of the players is 1.82 m with variance 0.0324 m2 . After checking the records again, three mistakes are found as shown below. Height of player recorded (m) 1.56 1.84 1.33 Actual height of player (m) 1.65 1.44 1.88 Calculate the correct mean and variance of the 20 football players. 21. For a frequency distribution, ∑f = 30, ∑f(x – x) 2 = 182.3, ∑fx2 = 1  025. Calculate the mean and standard deviation.

67 Mathematics Semester 3 STPM Chapter 1 Data Description 1 22. The age distribution of the population in a certain country is as follows. Age Number of people (in millions) 10 – 911 7.7 10 – 141 4.6 15 – 191 4.4 20 – 591 28.4 60 – 641 2.9 65 – 104 8.0 The interval 10 – 14 includes all people who are 10 years old and above but less than 15 years. (a) Display this distribution on a histogram. (b) Calculate an estimate of the median age. (c) Calculate an estimate of the mean age. 23. An experiment is carried out on the rearing of cows. The mass gain (in kilograms) for each of 100 cows during a certain period is recorded. The results are shown in the following table. Mass gain (kg) 5 – 9 10 – 14 15 – 19 20 – 24 25 – 29 30 – 34 Frequency 2 29 37 16 14 2 (a) Construct a histogram to represent the data. (b) Plot a cumulative frequency curve, and determine the median and interquartile range. (c) Calculate the mean and standard deviation. (d) State whether the statistics in (b) or (c) are better for representing the above data. Give a reason. 24. The table below shows the amounts of milk (in litres) obtained from 131 cows at a farm on a certain day. Amount of milk (litres) 5 – 10 11 – 16 17 – 22 23 – 28 29 – 34 35 – 40 Frequency 15 28 37 26 18 7 (a) Calculate the median and the mean amount of milk obtained. (b) State the shape of the distribution. Give a reason. 25. A study is made to determine the mean length of leaves taken from a type of tree. A sample of 100 leaves is chosen and their lengths measured to the nearest 0.1 cm. Mid-point of class (cm) 2.2 2.7 3.2 3.7 4.2 4.7 5.2 5.7 6.2 Frequency 3 5 8 12 18 24 20 8 2 (a) Draw and state the shape of the frequency curve for the lengths of the leaves. (b) Calculate an estimate of the mean using 4.7 cm as the assumed mean. (c) What are the boundaries of the class with mid-point 3.7 cm? (d) Construct a cumulative frequency table to estimate the median.

68 Mathematics Semester 3 STPM Chapter 1 Data Description 1 26. After checking a sample of 700 invoices, a marketing manager obtain the following frequency distribution. Amount on invoice (RM) Number of invoices 100 – 900 44 010 – 190 194 020 – 490 157 050 – 990 131 100 – 149 69 150 – 199 40 200 – 499 58 500 – 749 7 (a) Explain why the mean is not the best measure of central tendency of this distribution. (b) Calculate an alternative measure of central tendency which can be used by the marketing manager. 27. The total annual incomes of 100 families are summarised in the following table. Income (RMx) Number of families 6 000  x  10 000 2 10 000  x  12 000 40 12 000  x  14 000 37 14 000  x  18 000 11 18 000  x  24 000 8 24 000  x  30 000 2 (a) Display this frequency distribution on a histogram. (b) Calculate the Pearson coefficient of skewness. 28. 25 students estimated the length of a straight line in millimetres. The results are as follows: 42 76 36 53 22 11 34 3 88 41 7 48 9 40 14 13 25 10 26 18 26 30 34 45 38 (a) Determine the median and interquartile range. (b) Identify any outlier(s). (c) Draw a boxplot for this data. 29. The table below shows the lifespans of a sample of batteries. Lifespan (to the nearest hour) Frequency 620 – 639 4 640 – 659 8 660 – 679 16 680 – 699 37 700 – 719 40 720 – 739 34 740 – 759 20 760 – 779 15 780 – 799 12 (a) Calculate estimates of the median and interquartile range. (b) Draw a box-and-whisker plot.

69 Mathematics Semester 3 STPM Chapter 1 Data Description 1 30. The box-and-whisker plots for two distributions are shown below. 70 90 110 130 150 170 190 210 230 250 270 Group X Group Y Cents per day (a) Write down the median and interquartile range for each distribution. (b) State two differences between the two distributions. (c) State one similarity between the two distributions. 31. The following table shows the times, to the nearest minutes, spent reading during a particular day by a group of fifty students. Time (minutes) 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69 70 – 79 Number of children 4 10 17 9 6 3 1 (a) Construct the cumulative frequency distribution table for this distribution and plot the cumulative frequency curve. (b) Using the cumulative frequency curve, estimate the median and the interquartile range. (c) The reading time of one student is wrongly recorded as 26 minutes, while its actual value is 46. State whether this wrong value affects the median and interquartile range. Justify your answer. (d) Another two students are surveyed and both their reading times on a particular day are more than 81 minutes. (i) Out of 52 students, find the percentage of students with a reading time of less than 60 minutes. (ii) Calculate an estimate of the median reading times of the 52 students. 32. The mean and standard deviation of Chemistry marks for 28 school candidates and 7 private candidates are shown in the table below. Chemistry marks Number of candidates Mean Standard deviation School candidates 28 62 5 Private candidates 7 44 8 Calculate the overall mean and standard deviation of the Chemistry marks. 33. In an experiment to estimate the mean height, of eighteen-year-old boys, the heights, x cm, of 80 pupils were obtained. The data obtained were summarised by ∑(x – 160) = 250, ∑(x – 160)2 = 6888. Determine the mean and standard deviation of the boys.

70 Mathematics Semester 3 STPM Chapter 1 Data Description 1 34. The times, correct to the nearest second, for 100 athletes to run one lap of a running track were recorded in the following table. Recorded time (s) 71 – 75 76 – 80 81 – 85 86 – 90 91 – 95 96 – 100 Number of athletes 7 18 27 30 11 7 (a) Calculate the mean running time per lap. (b) Calculate the first quartile, median and third quartile of the athletes running times. (c) Estimate the percentage of athletes with a running time of less than 78 seconds. 35. A set of data has a mean of 6.4, a median of 5.6, and a standard deviation of 1.5. Calculate the Pearson coefficient of skewness and comment on your result. 36. In a survey, each respondent is required to rank the hygiene status of a hawker centre in town based on five aspects: surrounding, food, drinks, dishware and kitchen utensils. The ranking score ranges from 1 to 5 where 1 implies “extremely unhygienic” and 5 implies “extremely hygienic”. The average ranking score of each male and female respondent are summarised by the box-andwhisker plots below. 1.00 2.00 3.00 4.00 5.00 Males 0 Average ranking score Females Comment on the measure of central tendency, measure of dispersion and the skewness of the average ranking scores of the male and female respondents. 37. A lady has to take a bus from station A to her office every working day. The bus is scheduled to arrive at station A at 0800 hours every day. The actual time of arrival of the bus is recorded over a period of 80 working days and the results are shown in the following table. Time of arrival (t) Number of days 0754 , t < 0756 1 0756 , t < 0758 8 0758 , t < 0800 13 0800 , t < 0802 31 0802 , t < 0804 21 0804 , t < 0806 6 (a) If the man reaches the bus station at exactly 0755 hours every morning, calculate the average time, in minutes, he has to wait before the bus arrives for the 80 working days. (b) Construct a cumulative frequency table for the time of arrival of the bus. (c) Calculate the median and the mode for the time of arrival of the bus. (d) Calculate the semi-interquartile range for the time of arrival of the bus. (e) Determine the percentage of the days that the bus is late. 38. Twenty-three students estimated the length of a line. The results are shown below. 51 30 36 60 39 21 36 49 28 34 48 31 20 53 39 20 25 46 63 37 65 28 52 (a) Find the median and the quartiles of this distribution. (b) Use the quartiles to estimate the skewness of this distribution.

71 Mathematics Semester 3 STPM Chapter 2 Probability 2 CHAPTER PROBABILITY 2 Learning Outcome (a) Apply the addition principle and the multiplication principle. (b) Use the formulae for combinations and permutations in simple cases. (c) Identify a sample space, and calculate the probability of an event. (d) Identify complementary, exhaustive and mutually exclusive events. (e) Use the formula P(A  B) = P(A) + P(B) − P(A  B). (f) Calculate conditional probabilities, and identify independent events. (g) Use the formulae P(A  B) = P(A) × P(B | A) = P(B) × P(A | B). (h) Use the rule of total probability. addition principle – prinsip penambahan combination – gabungan complementary – pelengkap conditional probability – kebarangkalian bersyarat dependent – bersandar disjoint – tak bercantum equally likely – sama kemungkinannya event – peristiwa exhaustive event – peristiwa habisan independent – tak bersandar multiplication principle – prinsip pendaraban mutually exclusive – saling eksklusif outcome – hasil permutation – pilih atur relative frequency – kekerapan relatif sample space – ruang sampel total probability – jumlah kebarangkalian Bilingual Keywords

72 Mathematics Semester 3 STPM Chapter 2 Probability 2 2.1 Counting Techniques Addition principle of counting Let A1 , A2 , ..., Ak be disjoint events with n1 , n2 , …, nk possible outcomes, respectively. Then the total number of outcomes for the event “A1 or A2 or ... or Ak ” is n1 + n2 + … + nk . Suppose that we want to buy a fruit from one of two stalls A1 and A2 . Suppose also that those stalls have 10 and 15 different types of fruits, respectively. How many types of fruits are there altogether to choose from? Choosing one from given types of fruits from either stalls is called an event and the choices for either event are called the outcomes of the event. Thus the event “selecting one from stall A1 ”, for example, has 10 outcomes. Essentially, the addition principle says that if we want to count the number of ways either one case could happen or another case could happen, then we should add the number of ways each individual case could happen. Thus, we can choose one of 10 types of fruits from stall A1 or one of 15 types of fruits from stall A2 , there are altogether 10 + 15 = 25 types of fruits to choose from. Note that the events must be disjoint, that is they must not have common outcomes for this principle to be applicable. Example 1 Suppose there are 4 different flavours of noodle dishes and 7 different ingredients of fry rice dishes. How many selections does a customer have? Solution: An event is “selecting a dish of either kind”. There are 4 outcomes for the noodle event and 7 outcomes for the rice event. According to the addition principle, there are 4 + 7 = 11 possible selections. Multiplication principle of counting Let A1 , A2 , ..., Ak be events with n1 , n2 , ..., nk possible outcomes, respectively. Then the total number of outcomes for the sequence of these k events is n1 × n2 × … × nk . If we are buying a cup of ice cream that comes in a choice of three flavours from vanilla, chocolate or mango, and two sizes either small cup or large cup, how many different types of ice creams can be ordered? We have three choices for the flavours, for each choice of flavour; there are two choices of sizes. Selecting one of three choices is called an event, and a specific size is called the outcome of the event. The multiplication principle tells us that if we want to count the number of ways that one case could happen and another case could happen, then we should multiply the number of ways that each individual case could happen. Thus, we could order 3 × 2 = 6 different types of ice cream.

73 Mathematics Semester 3 STPM Chapter 2 Probability 2 Example 2 A password consists of three symbols. If the first symbol is a character, the second and the third symbols are digits, how many three-symbol passwords could be formed? Solution: Let A be the set of characters, B and C be the sets of digits. Then A × B × C is the set of all passwords fulfilled the requirement. There are 26 elements in the set of characters, 10 elements in the set of digits. We thus have n(A) × n(B) × n(C) = 26 × 10 × 10 = 2 600 Thus, a total of 2 600 three-symbol passwords could be formed. Example 3 Find the number of different 5 digit numbers. How many of these numbers are even? Solution: The first digit could be any numbers from 1 to 9. Each of the next four digits could be any digits. By the multiplication principle, there are 9 × 104 = 90 000 such numbers. Now the final digit must be one of the numbers 0, 2, 4, 6 or 8, i.e. 5 ways. Again by the multiplication principle, there are 9 × 103 × 5 = 45 000 such numbers. Note: If no letters can be duplicated in a label, then the first letter of a label can be selected from all 26 characters. The second letter, however, must be selected from 25 characters because one letter has been selected for the first position and that letter cannot be used for the second position. Similarly the third letter is now selected from the remaining 24 characters and the fourth from 23 characters. Permutations Assume that you try to arrange three persons A, B and C sitting in a row for dinner, how many ways can you line up the three persons? As the number is small, it is not difficult to make such arrangement. We could write down all the possibilities: ABC, ACB, BCA, BAC, CAB and CBA. But what if there were eight persons? Now, the arrangement is getting more complicated. We must come up a new counting method to handle this problem. The way to do this is to work in steps and then use the multiplication principle. Back to the case of three persons, at the initial stage we have three ways to fill the first seat, then two persons remain to fill the second seat and finally just one person left for the last seat. So, the number of ways to arrange the persons in order is 3 × 2 × 1 = 6. This multiplication principle works with any finite number of persons. For eight persons, the number of distinct arrangements is 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40 320. The above discussion introduces another basic counting principle. A permutation of a set of elements is a way of arranging all or part of the elements in a definite order. For n elements of a set to be orderly arranged, we can do this in n steps by line up one element at a time. There are n choices for the first place. Once the first place is filled, any one of the n – 1 elements can be filled in the second place, and so on. For each step, there is one less than the total elements in the previous step. By the multiplication principle, the number of permutations is n(n – 1)(n – 2) … 3 . 2 . 1

74 Mathematics Semester 3 STPM Chapter 2 Probability 2 Whenever we calculate permutation, product like this comes up very frequently. We represent this product by the notation n!, which is read “n factorial”. Thus, n! = n(n – 1)(n – 2) … 3 . 2 . 1 Note: By definition 1! = 1 and 0! = 1. First permutation rule The number of permutations of n distinct elements is n!. Example 4 There is a photo taking session in a birthday party. If 6 people line up for taking a photo, how many different ways can they be arranged from the left to the right? Solution: Any of the 6 people can be placed in the first position from the left. Once the first position is taken, there are 5 people left for the second position. After the first two positions are taken, there are only 4 people to choose for the third position, and so on. It is observed that there is one less person to choose from each time a position is taken. Thus, the number of ways the 6 people could be arranged is = 6 × 5 × 4 × 3 × 2 × 1 = 6! = 720 Thus, they can line up in any of the 720 possible ways. Sometimes, we may consider only a certain number of elements in a set to be arranged in order instead of all of them. For example, if you were to arrange ten students to sit in the first row but you only have three chairs, how many ways the chairs could be occupied? Although this problem is slightly different from previous examples, the approach in getting the solution is similar. Imagine there are three slots and the slots are to be filled one at a time. Any of 10 students may fill the first slot. After the first student is selected, any of 9 students may fill the second slot, and any of 8 students in the last slot. Possible ways: 10 9 8 The number of possible ways of placing 3 of the 10 students to sit in the first row is = 10 × 9 × 8 = 720 This product is commonly denoted by the symbol 10P3 . So, we have 10P3 = 10 × 9 × 8 = 720. 10P3 is read as “the number of permutations of 10 objects taken 3 at a time”. In general, n Prrepresents the number of ways r elements being selected from a set of n elements and placing them in order. By following the similar procedure as the above example, we have n Pr = n(n – 1)(n – 2) … [n – (r – 1)] = n(n – 1)(n – 2) … (n – r + 1) 1444442444443 r factors This expression can be simplified by multiplying (n – r)(n – r – 1) … 2 . 1 (n – r)(n – r – 1) … 2 . 1 , which is just equal to 1.

75 Mathematics Semester 3 STPM Chapter 2 Probability 2 n Pr = n(n – 1)(n – 2) … (n – r + 1) (n – r)(n – r – 1) … 2 . 1 (n – r)(n – r – 1) … 2 . 1 = n(n – 1)(n – 2) … (n – r + 1)(n – r)(n – r – 1) … 2 . 1 (n – r)(n – r – 1) … 2 . 1 = n! (n – r)! Second permutation rule The number of permutations of n distinct elements taken r at a time is n Pr = n! (n – r)! . Example 5 Evaluate each of the following permutations. (a) 8 P2 (b) 12P9 (c) 5 P5 (d) 6 P1 Solution: (a) 8 P2 = 8! (8 – 2)! = 8! 6! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 6 × 5 × 4 × 3 × 2 × 1 = 8 × 7 = 56 or 8 P2 = 8 × 7 = 56 by using multiplication principle. (b) 12P9 = 12! (12 – 9)! = 12! 3! = 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 3 × 2 × 1 = 79 833 600 (c) 5 P5 = 5! (5 – 5)! = 5! 0! 0! = 1 = 5! = 120 (d) 6 P1 = 6! 5! = 6 Note: 1. Example 5(c) demonstrates the first permutation rule, n Pn = n!. 2. Example 5(d) illustrates the case for selecting one element from a set of n elements, i.e. n P1 = n.

76 Mathematics Semester 3 STPM Chapter 2 Probability 2 Example 6 A badminton team has 5 players. How many ways can a coach select the first and second singles? Solution: Here we are choosing 2 from 5 and arranging them in order. So, the number of ways are 5 P2 = 5 × 4 = 20 Thus, there are 20 ways for the coach to select the first and second singles. So far we have considered permutations of distinct elements. If the letters B and C are both equal to X, then the 6 permutations of the letters A, B and C becomes AXX, AXX, XXA, XAX, XAX, XXA, of which only 3 are different. Thus, with 3 letters, 2 being the same, there are 3! 2! = 3 distinct permutations. In general, the number of distinct permutations of n elements of which n1 are of one kind, n2 of a second kind, …, nk of a kth kind is given by the formula n! n1 ! n2 ! … nk ! Example 7 Find the number of different ways to arrange 2 yellow, 3 red and 4 green bulbs in a string of Christmas tree lights with 9 sockets. Solution: The total number of different arrangements is 9! 2! 3! 4! = 1260 Example 8 Find the number of arrangements that can be formed from all the letters of the word MATHEMATICS, if (a) there is no restrictions, (b) the two M’s are separated. Solution: For the word MATHEMATICS, there are 11 letters with 2 A’s, 2 M’s, 2 T’s, 1 H, 1 E, 1 I, 1 C and 1 S. (a) If there is no restrictions, the total number of different arrangements = 11! 2! 2! 2! 1! 1! 1! 1! 1! = 4 989 600 (b) If the two M’s are together, they can be considered as a single letter and the total number of different arrangements = 10! 2! 2! 1! 1! 1! 1! 1! 1! = 907 200 Thus, if the two M’s are separated, the total number of different arrangements = 4 989 600 – 907 200 = 4 082 400 Note: This is an example of permutation with restriction.

77 Mathematics Semester 3 STPM Chapter 2 Probability 2 Combinations In dealing with permutations the order of the elements does matter. For instance, in Example 6 the coach’s selection of the first and second singles from the 5 players, the arrangement would be different if we exchanged the positions of the 2 selected players. Now, let us consider different type of selections. If the coach wished to pick a doubles from the 5 players, the order of choosing the first and the second player does not really matter. We begin with how many ways we can choose 2 players from a group of 5 players and arrange them in order. It was found to be 5 P2 = 5! 3! = 20 ways. Assume the players are designated by A, B, C, D or E. For a group of 2 players A and B, we have two arrangements, AB and BA. Note that the doubles consisting of AB is the same doubles formed by BA. This implies that within the 20 ordered arrangements, the order of the groups of 2 players should be disregarded. Consequently, we divide the ordered arrangements of 5 players taken 2 at a time, 5 P2 , by the number of ways of 2 players arranged between them, i.e. 2!. This gives 5 P2 2! = 5! 3! 2! = 5 × 4 2 = 10 doubles Generally, we could extend the above analysis by choosing r elements from a set of n elements without regard to order. This type of selections is classified as combination and is denoted by n Cr or 1 n r 2. The number of possible combinations, n Cr , is calculated by the second permutations formula divided by r!. Thus n Cr = n Pr r! = n! (n – r)! r! Example 9 How many different ways can you choose 4 kings from a standard deck of 52 playing cards? Solution: This problem involves combinations, because the order of the cards does not matter. We select 4 kings from 4 available kings. So, 4 C4 = 4! (4 – 4)! 4! = 1 There is only 1 way you could select 4 kings. Choosing without Return – A School Election VIDEO

78 Mathematics Semester 3 STPM Chapter 2 Probability 2 Example 10 From 5 men and 3 women, find the number of committees of 5 that can be formed with 3 men and 2 women. Solution: The number of ways of selecting 3 men from 5 is 5 C3 = 5! (5 – 3)! 3! = 10 The number of ways of selecting 2 women from 3 is 3 C2 = 3! (3 – 2)! 2! = 3 Thus, the required number of committees that can be formed = 10 × 3 = 30 Note: We apply the multiplication principle in this example. Note: A commonly raised question is, “When do we use permutation and when do we use combination?” To answer this we must first realise that objects are drawn from the lot without replacement. Thus, the total number of ways is n Pr if the order in which the objects are drawn is important, and n Cr if it is not important. Example 11 Consider the collection of objects consisting of the six letters a, b, c, d, e, f. Write the answers in factorial notations. (a) Find the number of three-letter permutations. (b) Find the number of three-letter combinations. (c) Which is greater, the number of permutations or the number of combinations? (d) Find an equation relating parts (a) and (b). Solution: (a) The number of three-letter permutations is, 6 P3 = 6! (6 – 3)! = 6! 3! (b) The number of three-letter combinations is, 6 C3 = 6! (6 – 3)! 3! = 6! 3! 3! (c) From parts (a) and (b), it is obvious that the number of permutations is greater than the number of combinations (d) The number of three-letter combinations is, 6 C3 = 6! 3! 3! = 6 P3 3! ⇒ 6 P3 = 6 C3 . 3! 6 P3 = 6 C3 . 3! is the required equation relating parts (a) and (b). In general, we can deduce that n Pr = n Cr . r!.

79 Mathematics Semester 3 STPM Chapter 2 Probability 2 Some useful tips for counting techniques When you use counting techniques, you must always keep these in mind: “Do I count everything?” and “Do I count anything more than once?” to avoid mistakes. It is not easy to answer the question which counting methods you should use. You may find some of the suggestions below helpful. 1. If the set of elements breaks up into disjoint subsets, then the addition rule can be applied. 2. If the elements are from different sets or the elements to be counted are collected through multiple steps, then the multiplication rule is used. 3. Permutations involve choosing a specific number of elements from a set of elements and placing them in order. 4. Combinations involve choosing a specific number of elements from a set of elements without regard to order. Exercise 2.1 1. There are two routes from a student’s home to a bus station and three routes from the bus station to the school. Use the multiplication principle to find the number of ways a journey from the student’s home to the school via the bus station may be completed. 2. If a Mathematics department schedules 4 lecture sections and 12 tutorial groups for a course in Introductory Statistics, in how many different ways can a student choose a lecture section and a tutorial group? 3. In a practical class, students are asked to classify the specimens according to their colours: red, blue, green, yellow, white and also according to their sizes: small, medium, large and lastly to their gender: male, female. In how many different ways can a specimen be classified according to colour, size and gender? 4. Jenny has 9 different blouses and 6 different skirts. Find the number of possible different blouse-skirt outfits she can form. If she also has 7 pairs of different stockings, how many different blouse-skirtand-stocking outfits are possible? 5. The license plates of some cars consist of two letters followed by three digits. Find the number of possible different license plates if (a) there is no restrictions, (b) no letter or digit is repeated. 6. How many lunches are possible consisting of soup, a sandwich, desserts, and a drink if one can select from 3 soups, 4 types of sandwiches, 4 desserts, and 5 drinks? 7. Find the number of ways for the letters of the word MONDAY to be arranged in a row. How many different ways can three of these letters be chosen and written in a row if (a) there is no restriction? (b) the first letter must be D? 8. How many ways can a football team schedule 3 exhibition games with 3 teams if they are all available on any of 5 possible dates? 9. How many ways can 8 people be assigned to 2 triple and 1 double rooms?

80 Mathematics Semester 3 STPM Chapter 2 Probability 2 10. A club has 30 members. They are to select 3 office holders consisting of the president, secretary, and treasurer for the following year. They always select these office holders by drawing 3 names randomly from names of all members. The first person selected becomes the president, the second is the secretary, and the third takes over as treasurer. Thus, the order in which 3 names are selected is important. Find the total arrangements of 3 names from 30. 11. Find the number of permutations that can be formed from the letters of the word POPULAR. How many of these permutations, (a) begin and end with P? (b) have the two P’s separated? (c) have the vowels together? 12. (a) In how many different ways can a student answers 8 true-false questions? (b) In how many ways may the test be completed if a student is imposed for each incorrect answer, so that the student may leave some questions unanswered? 13. An ice cream parlour has seven flavours of ice cream. Nathan wants to buy two flavours of ice cream. If he randomly selects two flavours out of seven, how many possible combinations are there? 14. Consider the collection of objects consisting of the five letters c, d, e, f, g. (a) List all possible combinations of three letters from this collection of five letters. (b) Use part (a) to determine the number of possible combinations of three letters that can be formed from this collection of five letters. (c) Use combination formula to confirm the answer obtained in part (b). 15. To recruit new members, a compact-disc (CD) club advertises a special introductory offer: A new member agrees to buy one CD at regular club prices and receives free any four CDs of the member’s choice from a collection of 50 CDs. Find the number of possibilities a new member has for the selection of the four free CDs. 16. In how many different ways can a coach choose two badminton players from among seven students and three table tennis players from among nine students? 17. Evaluate (a) 5 P5 (b) 8 P3 (c) 6 P4 (d) 12P0 (e) 7 C4 (f) 10C10 (g) 9 C5 (h) 15C0 18. Out of 5 class representatives and 7 prefects, a committee consisting of 2 class representatives and 3 prefects is to be formed. In how many ways can this be done if, (a) any class representative and any prefect can be included, (b) one particular prefect must be in the committee, and (c) two particular class representatives cannot be in the committee. 19. Santi has 6 flowers, each of a different variety. How many different bouquets can she form? 20. There are 25 paintings in a collection. How many ways can the following groups be selected? Write the answers in factorial notations. (a) 5 paintings, (b) 20 paintings What conclusion can you deduce from parts (a) and (b)? 21. A developer of a residential subdivision has four house styles and wants to build on eight adjacent lots. How many distinguishable arrangements are possible if the developer decides to build two houses of each style? 22. A school has 4 Mathematics teachers, 3 Chemistry teachers and 2 Biology teachers. Find the number of different sets of teachers a student could have for these 3 subjects.

81 Mathematics Semester 3 STPM Chapter 2 Probability 2 23. A company manager has to visit four of the ten subsidiaries that the company owns. (a) In how many different ways can the manager plan his itinerary in visiting four of the twelve subsidiaries? (b) How many sets of four companies are there from which the manager can pick one set to visit? 24. A company has 10 construction workers. The manager plans to assign 3 to job site A, 4 to job site B, and 5 to job site C. In how many different ways can the manager make this assignment? 2.2 Probability Sample spaces The scientists perform experiments to produce observations or measurements that will assist them in arriving at conclusions. In statistics the word experiment is used to describe any process that generates raw data or outcome. Thus, an outcome is a result of some activity. For example: Rolling a die has six outcomes: 1, 2, 3, 4, 5, 6 A sample space is a set of all possible outcomes for an activity and is represented by S. For example: The sample space for rolling a die is: S = {1, 2, 3, 4, 5, 6} The sample space for tossing a coin consists of the two outcomes, head H and tail T is: S = {H, T} An event is the collection of outcomes of particular interest in an experiment. For the experiment of tossing a coin, suppose we are interested only in getting a head, then this event consists of a single element, H. Thus, in this example of tossing a coin; The event that a head occurs is: E = {H}, If a coin is tossed 3 times in succession; The event that the number of heads is appears more than once is: E = {HHH, HTH, THH, HHT}, Obviously the event E is the subset of the sample space S. For another example in rolling a die; The event of obtaining a 3 is: E = {3}, The event of obtaining an odd number is: E = {1, 3, 5} Example 12 Consider the experiment of a team’s result in a football match. Determine the sample space, S. Solution: The possible outcome of the team’s result in a football match is win, lose or draw. So, the sample space is S = {win, lose, draw}. Example 13 For the experiment in Example 12, write the following events. (a) Event A: The team wins the match (b) Event B: The team does not lose the match Solution: (a) Event A has only one possible outcome, win, so the event A is: A = {win} (b) Two outcomes fulfil this given condition: win or draw. So the event B is: B = {win, draw}

82 Mathematics Semester 3 STPM Chapter 2 Probability 2 Example 14 A number is randomly picked from a set of integers, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. (a) Find the sample space of the above experiment. (b) List the outcomes in the following events for the above experiment. (i) The number is divisible by 3. (ii) The number is an even number. (iii) The number is 11. Note: By picking an integer at random, we mean that each integers has equally likely chance of being picked. Solution: (a) There are ten possible outcomes, so the sample space is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. (b) (i) The event of the number divisible by 3 has three possible outcomes, 3, 6 or 9. E = {3, 6, 9} (ii) The event of obtaining an even number has five possible outcomes, 2, 4, 6, 8, 10. E = {2, 4, 6, 8, 10} (iii) The set has ten integers and none of these integers is 11. So, picking a number 11 is impossible. We say it is an impossible event and E = {φ} Probability of an event If you are planning to visit your friend who is staying nearby, then you look at the sky and you are not sure whether it is going to rain. So, you are hesitating to bring along an umbrella with you. This is one of the non-deterministic situations that we encounter often. In the real world, many problems cannot be predicted with accuracy. Probabilities are thus introduced to deal with situations involving randomness or uncertainty about the outcome. Probability is a measure of how likely an event is to happen. In an experiment, if all the outcomes are equally likely to occur, then the probability of an event to occur is the number of ways that an event can occur divided by the total number of outcomes in the sample space. P(Event) = number of ways that an event can occur total number of possible outcomes The probability of an event is a number between 0 and 1 that indicates the likelihood the event will occur as shown below: P(E) = 0 P(E) = 0.5 P(E) = 1 Note: 1. P(E) = 0 means that the event will not occur. 2. P(E) = 1 means that the event is certain to occur. 3. P(E) = 0.5 means that the event is equally likely to occur or not occur. 4. The closer the probability of a given event is to 1, the more likely it is to occur. 5. Probability can be expressed as decimal, fraction, ratio or percentage. 6. For example: P(E) = 0.5, 1 2 or 50%.

83 Mathematics Semester 3 STPM Chapter 2 Probability 2 Definition: 1. The probability that an event will happen is between 0 and 1 inclusive, i.e. 0 < P(E) < 1. 2. P(φ) = 0; φ is 0 collections. 3. P(S) = 1 Example 15 A fair die is tossed. If the event of interest is obtaining a number less than 3, find the probability of the event happening. Solution: There are six different possible outcomes in the sample space for tossing a die, so S = {1, 2, 3, 4, 5, 6}. Let E be the event of obtaining a number less than 3. Event E has two outcomes satisfying the requirement, we thus have E = {1, 2}. We use a Venn diagram to show the relationship between S and E. S 1 E 3 4 5 6 2 The probability of the event E occurring is P(E) = 2 6 = 1 3 Thus, the probability of getting a number less than 3 is 1 3 . Definition: If an experiment can result in any one of N different equally likely outcomes, and if exactly n of these outcomes correspond to event E, then the probability of event E is P(E) = n N Example 16 Two fair coins are tossed. (a) Use a tree diagram to find the sample space of the above experiment. (b) List the simple events in the sample space and their corresponding probabilities. (c) Find the probability of observing exactly one head in the two tosses. Solution: (a) Let H and T denote head or tail respectively. The tree diagram is as follows: First coin Outcome H H Second coin HH HT TH TT T H T T The sample space of the above experiment is {HH, HT, TH, TT}

84 Mathematics Semester 3 STPM Chapter 2 Probability 2 (b) The simple events in the sample space and their corresponding probabilities are listed in the following table: Event First coin Second coin P(E) E1 H H 1 4 E2 H T 1 4 E3 T H 1 4 E4 T T 1 4 (c) Let F = event of observing exactly one head in the two tosses. The probability of observing exactly one head in the two tosses is, F = {HT, TH} P(F) = 2 4 = 1 2 Example 17 3 students are chosen from a group of 10 students consisting of 7 boys and 3 girls to represent a school in chess competition. If the selection is merely based on random picking, find the probability that the representatives are formed by (a) 2 boys and 1 girl, (b) at least 2 girls. Solution: (a) There are 10C3 = 10! (10 – 3)! 3! = 10 × 9 × 8 3 × 2 × 1 = 120 combinations by taking 3 from 10 students. There are 7 C2 × 3 C1 = 7 × 6 2 × 1 × 3 1 = 63 combinations by exactly taking 2 boys and 1 girl from the group. Hence, the probability that the representatives are formed by exactly 2 boys and 1 girl is 63 120 = 21 40 . (b) The selection could be either 2 girls and 1 boy or 3 girls and no boys. So, the number of combinations of 2 girls and 1 boy or 3 girls and no boys = 3 C2 × 7 C1 + 3 C3 × 7 C0 = 3 × 7 + 1 × 1 = 22 The probability that the representatives are formed by at least 2 girls is 22 120 = 11 60 .

85 Mathematics Semester 3 STPM Chapter 2 Probability 2 In summary, if the number of equally probable outcomes in the sample space S is denoted by n(S) and the number of equally probable outcomes in an event E is written as n(E), then the probability of an event to occur can be expressed as P(E) = n(E) n(S) . As E is a subset of S, we have 0 < n(E) < n(S). Dividing all by n(S), 0 n(S) < n(E) n(S) < n(S) n(S) Thus, we get 0 < P(E) < 1. Another approach in determining the probability is based on relative frequency. If an experiment is repeated n times under the identical condition and an event is observed to happen f times, the probability of the event happening is then estimated to be P(E) = frequency of the event occured total number of observations = f n Example 18 Data are collected on the gender of customers who enter a supermarket on a particular day. It is found that out of 360 customers, 249 are females. Find the probability that a customer who visits the supermarket on that day is a female customer. Solution: Let A be the event that a female customer visits the supermarket. P(A) = 249 360 = 83 120 Complementary events Two events are said to be complementary, if one event happens then the other event cannot happen at the same time. Both events contain all the experimental outcomes in the sample space. Let E (read as E prime) denotes the event E does not happen where E is called the complement of E. If n(S) is the size of the sample space, n(E) is the number of outcomes in event E, then n(E) = n(S) – n(E). Hence, in terms of probability P(E) + P(E) = n(E) n(S) + n(E) n(S) = n(E) + n(S) – n(E) n(S) = n(S) n(S) = 1 This rule for complementary events states that if two events are complementary, then the sum of their probabilities equal to 1. Hence, P(event E happens) + P(event E does not happen) = 1. Rearranging this equation we obtain the complement rule as follows: P(E) = 1 – P(E)

86 Mathematics Semester 3 STPM Chapter 2 Probability 2 Example 19 A die is tossed. (a) Calculate the probability of, (i) obtaining an odd number, (ii) getting an even number. (b) Show the validity of the rule for complementary events. Solution: (a) The sample space for tossing a die is S = {1, 2, 3, 4, 5, 6}. (i) Let A be the event of obtaining an odd number. Event A = {1, 3, 5} P(A) = 3 6 = 1 2 (ii) Let B be the event of getting an even number. Event B = {2, 4, 6}. P(B) = 3 6 = 1 2 (b) P(A) + P(B) = 1 2 + 1 2 = 1 We have thus illustrated the rule of complementary events. Example 20 A letter is randomly selected from the alphabet. Find the probability of not getting a vowel. Solution: Let A be the event of getting a vowel. There are 5 vowels {a, e, i, o, u} and a total of 26 letters in the alphabet. Thus, P(A) = n(A) n(S) = 5 26 . The event of not getting a vowel is the complement of the event A. By the rule of complementary events, we have P(A) = 1 – P(A) = 1 – 5 26 = 21 26 Alternatively, we can find the probability directly by counting the letters that are not vowels. So, n(A) = 21 and P(A) = 21 26 . Note: Sometimes using complementary events can make the probability calculation easier.

87 Mathematics Semester 3 STPM Chapter 2 Probability 2 Exhaustive events Two events are said to be exhaustive if it is certain that at least one of them occurs. If the events A and B are exhaustive, they together form the whole sample space. In set language, A  B = S. For example, when tossing a die the events “getting an even number” and “getting an odd number” are exhaustive, because they include all possible outcomes. Example 21 In a group of 10 students, 5 are form-one boys, 3 are form-one girls and the remaining 2 are form-two girls. A student is randomly chosen from the group. The events A, B, C and D are defined as follows. A : The selected student is a form-one student, B : The selected student is a girl, C : The selected student is a boy, D : The selected student is a form-two girl, Identify which pairs of the events are exhaustive. Solution: The following pairs of events are exhaustive. • A and B, because it is certain that at least one of both events occur. • A and D, because it is certain that at least one of both events occur. • B and C, because it is certain that at least one of both events occur. If the events A and B are exhaustive, then A  B = S. So, P(A  B) = 1. Mutually exclusive events We are often interested in finding the probability of events whose outcomes are described by two or more other events. For example, a secondary school data shows that 28% of students age 13 years and 16% age 17 years. If a student from the school is selected at random, what is the probability that the student ages 13 years or 17 years? In the following section we expand the probability calculation to include two or more events. Two or more events are mutually exclusive or disjoint if the events cannot occur at the same time when the experiment is performed. Mutually exclusive events can be shown by using a Venn diagram as follows: S A B If A and B are two mutually exclusive events, then (a) they do not have any outcomes in common or cannot both occur at the same time, i.e. A  B = φ, the intersection of A and B is the empty set. (b) P(A and B) = 0, i.e. P(A  B) = 0 (c) P(A or B) = P(A) + P(B), i.e. P(A  B) = P(A) + P(B). Demonstrating Exclusive Events VIDEO

88 Mathematics Semester 3 STPM Chapter 2 Probability 2 Note: If events A and B are not only exhaustive but they are also mutually exclusive, then exactly one of the events can happen. For instance, an event and its complement are always exhaustive and mutually exclusive: P(A  A) = 1 as well as P(A  A) = 0. Example 22 A fair die is tossed, let event A = obtain a one, event B = obtain a four, event C = obtain an even number, event D = obtain an odd number. Which of the following groups of events are mutually exclusive? (a) A and B (b) B and C (c) B and D Solution: (a) Events A and B are mutually exclusive because a single die can only land one way. Obtaining both “one” and “four” at the same time is impossible. (b) Events B and C are not mutually exclusive because they have the common outcome “four”. Both events occur if the number “four” is obtained. (c) Events B and D are mutually exclusive. As shown in the Venn diagram, there is no overlapping between the two events. It is impossible to get a number that is both “four” and an odd number in a single toss. 2 1 3 5 6 4 B A C D Example 23 A coin is tossed. (a) List the possible outcomes. (b) Define the sample space. (c) List the simple events. (d) Are the events mutually exclusive? (e) Are the events exhaustive? (f) Assuming that there is equal probability for the coin to land with any of its faces up and that it will not stand on its edge, find the probability of each event. Solution: (a) The possible outcomes are head and tail. Let H represents head and T represents tail. (b) The sample space, S = {H, T}. (c) The simple events are {H} and {T}. Let E1 = {H} and E2 = {T}. (d) Since on any toss, either H or T may turn up, but not both; the events are mutually exclusive.

89 Mathematics Semester 3 STPM Chapter 2 Probability 2 (e) Since the events E1 and E2 make up the whole of the possibility space, hence, E1 and E2 are exhaustive events. (f) For exhaustive events E1 and E2 , P(E1  E2 ) = 1 For equally probable outcomes P(E1 ) = P(E2 ), P(E1 ) + P(E2 ) = 1 2P(E1 ) = 1 P(E1 ) = 1 2 = P(E2 ) Probability of the union of events In Example 22 we conducted an experiment by tossing a die. Event B was “obtaining a four” and event D was “obtaining an odd number”. The probability of event B, P(B), is 1 6 and the probability of event D, P(D), is 3 6 or 1 2 . Suppose we want to find the probability of tossing an odd number or four, which is denoted by P(B  D). We notice that events B and D are mutually exclusive. Hence, P(B  D) = n(B  D) n(S) = n(B) + n(D) n(S) = n(B) n(S) + n(D) n(S) = P(B) + P(D) = 1 6 + 1 2 = 2 3 Example 24 It is found that out of 100 students of a school, 65 students go to school by bus, 15 students walk to school and the remaining 20 students go to school by other transportations. If a student is randomly selected from this group of 100 students, what is the probability that the student selected either goes to school by bus or walks to school? Solution: Let event B be “the student selected goes to school by bus” and event W be “the student selected walks to school”. The two events are mutually exclusive as the selected student cannot go to school by two different means at the same time. Thus, we have P(B  W) = P(B) + P(W) = 65 100 + 15 100 = 4 5 If two events A and B are mutually exclusive, the probability of A or B occurring is P(A  B) = P(A) + P(B) This is the addition rule for mutually exclusive events:

90 Mathematics Semester 3 STPM Chapter 2 Probability 2 The events linked to an experiment may not be mutually exclusive. Consider the following case: Suppose 10 students were asked what sports they have participated recently, their answers showed that 5 students played football and 7 students played badminton. What is the probability that one of the students selected has played football or badminton? The probability of selecting a student who has played football is 5 10 or 0.5 and the probability of a student playing badminton is 7 10 or 0.7. If the addition law for mutually exclusive events is used, the sum of these two probabilities is 1.2. We know that the value of probability cannot exceed 1. So, let us check the counting. To count all the students playing either football or badminton, we need to count all the students playing football, all the students playing badminton and subtract from this the number of students who were counted twice because they were playing both football and badminton. To help us visualise the above reasoning, the Venn diagram is used. Assume that 3 students played both football and badminton in the above example. The diagram indicates clearly why 3 outcomes in the overlap area of the event A  B are being counted twice – once in A and once in B. Where event A represents “a selected student playing football” and event B represents “a selected student playing badminton”. A B For events that are not mutually exclusive, the addition law is modified to take into account of double counting. If the number of outcomes in event A is n(A) and the number of outcomes in event B is n(B), then to find the number of outcomes in event A  B, we must count the outcomes in A  B. So, to get the correct total without counting the outcomes in the overlap A  B twice, we must subtract the number of outcomes in A  B. Thus n(A  B) = n(A) + n(B) – n(A  B) Let S be a sample space with n(S) possible outcomes. The probability of event A  B is given by P(A  B) = n(A  B) n(S) = n(A) + n(B) – n(A  B) n(S) = n(A) n(S) + n(B) n(S) – n(A  B) n(S) = P(A) + P(B) – P(A  B) This is called addition rule of probability: P(A  B) = P(A) + P(B) – P(A  B) A S B

91 Mathematics Semester 3 STPM Chapter 2 Probability 2 Example 25 Consider the experiment of selecting one card at random from a standard deck of 52 playing cards. Find the probability of drawing either a king or a diamond card. Solution: Let event A = a king card is drawn, event B = a diamond card is drawn. As there are 4 king cards in the deck, P(A) = 4 52 = 1 13 , and the deck has 13 diamond cards, so P(B) = 13 52 = 1 4 . Since there is 1 card corresponding the king of diamond, P(A  B) = 1 52 . By applying the formula, P(A  B) = P(A) + P(B) – P(A  B), we have P(A  B) = 1 13 + 1 4 – 1 52 = 4 13 The probability of drawing either a king or a diamond card is 4 13 . Relative frequency data for two or more events is often summarised in a table called a contingency table. We can easily determine probabilities from this table. Example 26 A survey of 150 students on their reading habit over the weekend is presented in the contingency table. Newspaper Total Yes No Magazine Yes 16 21 37 No 75 38 113 Total 91 59 150 If a student under the survey is selected at random, find the probability that the student reads newspaper or reads a magazine. Solution: Let A be the event that the student selected reads newspaper, B be the event that the student selected reads a magazine. From the table, there are 91 students out of a total of 150 students reading newspapers, hence P(A) = 91 150 . There are 37 students out of a total of 150 students reading magazine, hence P(B) = 37 150 . The probability that the selected student reads both newspaper and a magazine, P(A  B) = 16 150 .

92 Mathematics Semester 3 STPM Chapter 2 Probability 2 Using the formula, P(A  B) = P(A) + P(B) – P(A  B), we get P(A  B) = 91 150 + 37 150 – 16 150 = 56 75 The probability that the selected student reads newspaper or a magazine is 56 75 . Conditional probability A box contains 6 red and 4 green balls. Suppose two balls are chosen at random from this box. Let us define two events, A = the first ball is red and B = the second ball is red. Consider the situation that the ball is chosen without replacement. The calculation of the probability that the first ball chosen is red P(A) is straight forward and it is found to be 6 10 = 3 5 . What about the probability of choosing the second red ball P(B)? To compute P(B) we need to know whether the event that the first ball drawn is red or did not happen. This example introduces an important concept called conditional probability. In many circumstances the probability of an event is affected by the occurrence of another event. Conditional probability is defined as the probability of event A happening given that event B has happened. It is denoted by P(A | B). Note: Conditional events: If A and B are two events such that A will occur given that B has already occurred, then A given B or A | B is the event A conditional on B happening. The vertical bar “| ” is read as “given that”. Consider the following Venn diagram. Suppose that event B is chosen and we wish to find the probability that event A is being picked. Now the sample space is reduced from the original sample space S to B with this additional information. A S B Thus, to compute the conditional probability of the event A given that event B occurred, denoted by P(A | B), we need to count the numbers of elements in event A  B as well as in event B. We could write P(A | B) = n(A  B) n(B) . Divide both numerator and denominator by the total number of elements in the sample space n(S) to get P(A | B) = n(A  B) n(S) n(B) n(S) = P(A  B) P(B) .

93 Mathematics Semester 3 STPM Chapter 2 Probability 2 Hence, the conditional probability can be determined using the following formula. P(A | B) = P(A  B) P(B) , provided that P(B) ≠ 0 Rearranging the above equation gives P(A  B) = P(A | B) × P(B) Note: 1. P(B  A) = P(B | A) × P(A), since P(A  B) = P(B  A) we deduce that, P(A | B) × P(B) = P(B | A) × P(A) 2. If A and B are mutually exclusive events, i.e. A  B = φ, then P(A  B) = 0. Thus, P(A | B) × P(B) = P(B | A) × P(A) = 0. 3. For mutually exclusive and exhaustive events A and A, A  A = φ, A  A = S, P(A  A | B) = P(A | B) + P(A | B) i.e. P(S | B) = P(A | B) + P(A | B). P(S | B) = P(S  B) P(B) = P(B) P(B) = 1 Hence, P(A | B) + P(A | B) = 1 or P(A | B) = 1 – P(A | B) 4. Consider the following Venn diagram: A A fi Bfi A fi B S B A = (A  B)  (A  B) P(A) = P(A  B) + P(A  B)  (A  B) and (A  B) are mutually exclusive. ⇒ P(A) = P(A | B) . P(B) + P(A | B) . P(B) Example 27 A bag contains 8 yellow and 4 green marbles. Two marbles are taken randomly from the bag. Determine the probability that only one of them is yellow. Solution: Let event A = first marble selected is yellow, event B = second marble selected is yellow. Then, P(A) = 8 12 = 2 3 . When the second marble is selected, there are only 11 marbles left and 4 green marbles remain untouched. So the conditional probability of B given that A has happened is:

94 Mathematics Semester 3 STPM Chapter 2 Probability 2 P(B | A) = 4 11 . By applying the conditional probability formula, P(yellow first and green second) = P(A  B) = P(B | A) × P(A) = 4 11 × 2 3 = 8 33 Similarly, P(A) = 4 12 = 1 3 . When the second marble is selected, there are only 11 marbles left and 8 yellow marbles are still there. The conditional probability of B given that A has occurred is thus determined and P(B | A) is equal to 8 11 . Using the conditional probability formula, P(green first and yellow second) = P(A  B) = P(B | A) × P(A) = 8 11 × 1 3 = 8 33 . P(only one yellow) = P(first yellow followed by second green or first green followed by second yellow) = P(A  B) + P(A  B) = 8 33 + 8 33 = 16 33 . The probability that only one of the two marbles taken is yellow is 16 33 . Note: The result whether the two marbles are taken from the bag at the same time or one after the other is the same. Example 28 A and B are two events in a sample space S such that: P(A | B) = 0.5, P(A) = 0.65 and P(B) = 0.7. Calculate (a) P(B | A), (b) P(B | A). Solution: (a) P(B) = 1 – P(B) = 1 – 0.7 = 0.3 P(A | B) = P(A  B) P(B) 0.5 = P(A  B) 0.3 P(A  B) = 0.15 P(B | A) = P(A  B) P(A) = 0.15 0.65 = 3 13

95 Mathematics Semester 3 STPM Chapter 2 Probability 2 (b) P(A) = 1 – P(A) = 1 – 0.65 = 0.35 P(A  B) = P(B) – P(A  B) = 0.3 – 0.15 = 0.15 P(B | A) = P(A  B) P(A) = 0.15 0.35 = 3 7 Example 29 In a factory, three machines A, B and C are operated to make certain parts. The percentages of the parts manufactured by the machines A, B and C are 35%, 50% and 15% respectively. It is known that 8%, 5% and 16% of the parts produced by the machines A, B and C respectively are defective. If a finished part is randomly picked, calculate the probability that the part is from the machine A given that it is defective. Solution: Let event J = a part is made by machine A, event K = a part is defective. Assume that the total parts manufactured by the three machines are n. The numbers of parts produced by machines A, B and C are 0.35n, 0.5n and 0.15n respectively. Thus, the defective parts produced by machines A, B and C are 0.08 × 0.35n, 0.05 × 0.5n and 0.16 × 0.15n respectively. Thus, P(K) = 0.08 × 0.35n + 0.05 × 0.5n + 0.16 × 0.15n n = 0.077 and P(J  K) = 0.08 × 0.35n n = 0.028 Applying the conditional probability formula, the probability of the part made by machine A given that the part is defective, P(J | K) = P(J  K) P(K) = 0.028 0.077 = 0.364 Independent events Suppose a fair coin is tossed and a ‘head’ is shown face up. What would the coin land on for the next toss? The probability of getting a ‘head’ or a ‘tail’ is still 0.5. The outcome of the second toss is not affected by the previous result. When the knowledge that an event has happened provides no information about the occurrence of another event, the two events are said to be independent. Thus, if the outcome of event A does not affect the outcome of event B, then A and B are independent events, i.e. P(A | B) = P(A) or, equivalently P(B | A) = P(B) A Afi fi B S B