Mathematics SM025 Smart Revision Slot 3

‘Smart Revision Slot 3’

LIM HWEE CHENG
Kolej Matrikulasi JOHoR

Question 1

Lily buys 5 jam doughnuts, 4 cream doughnuts and 3 plain

doughnuts. On arrival home, each of her three children eats

one the twelve doughnuts. The different kinds of doughnut are

indistinguishable by sight and so selection of doughnuts is

random. Calculate the probability of the following events.

(a) All 3 doughnuts eaten contain jam. [2 marks]

(b) All 3 doughnuts are of the same kind. [2 marks]

(c) The 3 doughnuts are all of a different kind. [2 marks]

Suggested answer scheme

5 jam doughnuts Total = 12
4 cream doughnuts
3 plain doughnuts.

a) P(all jam) = 5 C3 K1
12 C3 J1

=1
22

 0.0455

Suggested answer scheme

5 jam doughnuts Total = 12
4 cream doughnuts
3 plain doughnuts.

b) P(all same kind ) = 5C3 + 4C3 + 3C3

12C3 12C3 12C3
=3

44
 0.0682

K1

J1

Suggested answer scheme

5 jam doughnuts Total = 12
4 cream doughnuts
3 plain doughnuts.

K1

c) P(all different ) = 5 C1 4 C1 3 C1
= P(JCP) + P(JPC) + P(CJP) + 12 C3
P(CPJ) + P(PJC) + P(PCJ)
=3
11

 0.2727

J1

P( X = x) = a(3 − x), x = 0, 1, 2
 x=3
 b, Question 2

The discrete random variable X has probability function

P( X = x) = a(3 − x), x = 0, 1, 2
 x=3
 b,

(a) Given that E(X) =1.6. Find the values of a and b.

[4 marks]

(b) Find E(3X – 2) [2 marks]

Suggested answer scheme

a) E(X ) =  xf (x) = 1.6 K1

0 + a(2) + 2a(1) + 3b = 1.6

4a + 3b = 1.6 ………………..

 f (x) = 1  3a + 2a + a + b = 1 K1

6a + b = 1 ……………

Solving  K1
J1
a = 0.1 b = 0.4

Suggested answer scheme

b) E(3X − 2) = 3E(X ) − 2 K1
= 3(1.6) – 2

= 2.8

J1

Question 3

Any patients who fails to turn up for an outpatient appointment
at a hospital is described as a ‘no-show’. At a particular
hospital, on average 15% of patients are no-shows.
(a) A random sample of 20 patients who have outpatient

appointments is selected. Find the probability that
(i) there is exactly 1 no-show in the sample.

[2 marks]
(ii) there are at most 2 no-shows in the sample.

[2 marks]

Suggested answer scheme

a) X ~ B(20, 0.15)

i) P(X = 1) = P(X ≥ 1) − P(X ≥ 2)

= 0.9612 – 0.8244 K1

= 0.1368 J1

i) P(X ≤ 2) = 1 – P(X ≥ 3) K1
= 1 – 0.5951

= 0.4049 J1

Question 3

(b) Exactly 100 patients who have outpatient

appointments is selected. By using a suitable

approximation,

(i) find the probability that between 20 and 25

(inclusive) no-shows in the sample.

[5 marks]

(ii) If the probability that at most p no shows is

0.2420, find the value of p. [4 marks]

Suggested answer scheme

b) n = 100 p = 0.15 np = 15 npq = 12.75

X ~ N(15, 12.75) B1 B1
i) P(20 ≤ X ≤ 25)
K1 Continuity
= P191.52−.7155  Z  25.5 − 15  correction &
12.75 standardizing

= P(1.26 < Z < 2.94) K1

= P(Z > 1.26) – P(Z > 2.94)

= 0.1038 – 0.00164

= 0.10216 J1 1.26
2.94

Suggested answer scheme

b) n = 100 p = 0.15 np = 15 npq = 12.75

ii) P(X ≤ p) = 0.2420

P Z  (p + 0.5) − 15  = 0.2420 K1 Continuity
 12.75  correction &
standardizing

0.2420 (p + 0.5) − 15 = −0.7 J1 – 0.7
12.75

(p + 0.5) − 15 p = (−0.7  12.75) +15 − 0.5 Solving
p = 12 J1 K1 for p
12.75

– ve value

Question 4

A disease is known to be present in 2% of a population. A
test is developed to help determine whether or not someone
has the disease. If the person has the disease, the test is
positive with probability 0.95. Whereas, if the person does
not have the disease, the test is positive with probability
0.03.

Suggested answer scheme

(a) Draw a tree diagram to represent the information.
[3 marks]

0.95 Correct
0.05
+ve K1 branches

0.02 Disease
0.98
– ve Correct prob

No 0.03 J1 (0.02, 0.95 &
Disease 0.97 0.03)

+ve

– ve J1 All correct

Suggested answer scheme

(b) A person is selected at random from the population
and tested for the disease. Find the probability that
the test is negative.
[2 marks]

P(negative) = 0.02(0.05) + 0.98(0.97) K1

= 0.9516 J1

Suggested answer scheme

(c) A doctor randomly selects a person from the

population and tests him for the disease. Given that

the test is positive. Find the probability that he does

not have the disease.

P(no disease | + ve) [3 marks]

= P(no disease  +ve) K1 numerator
P(+ve) K1 denominator

= 0.98(0.03)
0.02(0.95) + 0.98(0.03)

= 0.6074 J1

Suggested answer scheme

(d) If two persons are selected at random and tested for
the disease, find the probability that
(i) both of them test positive.
[2 marks]

P(+ve  +ve) = (0.484)2 K1

= 0.00234

J1

Suggested answer scheme

(d) If two persons are selected at random and tested for
the disease, find the probability that
(ii) at least one of them tests negative.
[3 marks]

P(–ve  +ve) or P(+ve  –ve) or P(–ve  –ve)

= (0.9516 × 0.0484) + (0.0484 × 0.9516) + (0.9516)2

= 0.9977 K1 K1
All correct
J1 At least 1
correct

a=1 Question 5
2
. [3 marks]

The cumulative distribution function of X of a continuous
random variable is given by

0, x  0

 1 x, 0 x 1

2
F ( x) =  x − a, 1 x  3
 2


 x 3
2
b,

Suggested answer scheme

(a) State the value of b and show that a = ½ .

[3 marks]

b=1 B1
3−a=1 K1
2 J1

a= 1
2

Suggested answer scheme

(b) Find P 1 X  5  [2 marks]
 2 4 
K1
= F 5  − F 1 
 4   2 

=  5 − 1  −  11 
 4 2   22 

= 1 J1
2

Suggested answer scheme

(c) Find P  X  6  [2 marks]
 5 

= 1− P  X  6 
 5 

= 1− F 6 
 5 

= 1−  6 − 1  K1
5 2

=3 J1
10

Suggested answer scheme

d) Find the probability density function of X.

[3 marks]

d  1 x  = 1 d  x − 1  =1
dx  2  2 dx  2 

K1 K1

 21 , 0 x1


 3 J1
f (x) =  1, 1 x  2



 0, otherwise


Suggested answer scheme

(e) Find the expected value of X.

[3 marks]

1.5 K1

E(X ) =  xf (x) dx 1  9 21 
4  8
0

01 x 1.5 = + −

= 1  dx +  x(1) dx
2
1 =7
8
J1

 x2  1  x2  1.5
  
= +  2  1 K1

 4  0

a=1 Question 6
2
. [3 marks]

(a) Given that P(J) = 0.25, P(J  K) = 0.15 and

P(J ’  K ’) = 0.7.

(i) Find P(J  K). [2 marks]

P(J  K) = 1 – P(J ’  K ’)

= 1 – 0.7 K1

= 0.3 J1

a=1 Question 6
2
. [3 marks]

(a) Given that P(J) = 0.25, P(J  K) = 0.15 and
P(J ’  K ’) = 0.7.

(ii) Find P(K). [2 marks]

P(J  K) = P(J) + P(K) – P(J  K)
0.3 = 0.25 + P(K) – 0.15 K1

P(K) = 0.2 J1

a=1 Question 6
2
. [3 marks]

(a) Given that P(J) = 0.25, P(J  K) = 0.15 and
P(J ’  K ’) = 0.7.

(iii) Find P(K | J). [2 marks]

P(K | J) = P(K  J)
P(J)

= 0.15 K1
0.25

= 0.6 J1

a=1 Question 6
2
. [3 marks]

(a) Given that P(J) = 0.25, P(J  K) = 0.15 and
P(J ’  K ’) = 0.7.

(iv) Determine whether or not J and K are

independent events. [3 marks]

?

P(K  J) = P(K)  P(J)

? K1 J1
0.15 = 0.2 × 0.25
 J & K are not
0.15 ≠ 0.05
J1 independent events

a=1 Question 6
2
. [3 marks]

b) In a village, power cuts occur randomly at a rate of 3

per year

(i) Find the probability that in any given year there

will be exactly 7 power cuts, [3 marks]

X = the number of power cuts.

X ~ Po(3) B1 K1
(i) P(X = 7) = P(X ≥ 7) – P(X ≥ 8)

= 0.0335 – 0.0119

= 0.0216 J1

a=1 Question 6
2
. [3 marks]

b) In a village, power cuts occur randomly at a rate of 3

per year

(i) Find the probability that in any given year there

will be exactly 7 power cuts, [3 marks]

X = the number of power cuts.

X ~ Po(3) B1 Alternative
(i) P(X = 7) = e−337 solution

7! K1

= 0.0216 J1

a=1 Question 6
2
. [3 marks]

b) In a village, power cuts occur randomly at a rate of 3 per

year.

(ii) Find the probability that there will be between 15 to

25 (inclusive) power cuts in 10 years time.

[3 marks]

X = the number of power cuts.

X ~ Po(30) B1

ii) P(15 ≤ X ≤ 25) = P(X ≥ 15) – P(X ≥ 26)

= 0.9991 – 0.7916 K1

= 0.2075 J1

Prepared by: LIM HWEE CHENG