GRADE 10 HOTS -PROBABILITY

Grade 10 PROBABILITY Mathematics

Question 1) Find the probability that a leap year selected
at random, will contain 53 Mondays.
Solution:
In a leap year, total number of days = 366
∴ 366 days
= 52 complete weeks + 2 extra days
Thus, a leap year always has 52 Mondays and extra 2 days.
Extra 2 days can be,
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
Let E be the event that a leap year has 53 Mondays.
∴ E = {Sun and Mon, Mon and Tues}
∴ P(E) = 2/7

Question 2) A bag contains cards numbered from 1 to 49. A

card is drawn from the bag at random, after mixing the cards

thoroughly. Find the probability that the number on the drawn
card is:
(i) an odd number
(ii) a multiple of 5
(iii) a perfect square
(iv) an even prime number
Solution:
Total number of cards = 49
(i) Odd numbers are 1, 3, 5, …., 49, i.e., 25
∴ P(an odd number) = 25/49

(ii) ‘A multiple of 5’ numbers are 5, 10, 15, ……., 45, i.e., 9
∴ P(a multiple of 5) = 9/49

(iii) “A perfect square” numbers are 1, 4, 9, …….., 49, i.e., 7
∴ P(a perfect square number) = 7/49=1/7

(iv) “An even prime number” is 2, i.e., only one number
∴ P(an even prime number) = 1/49

Question 3) A box contains 20 cards numbered from 1 to 20. A

card is drawn at random from the box. Find the probability that
the number on the drawn card is (2015D)
(i) divisible by 2 or 3,
(ii) a prime number.
Solution:
(i) Numbers divisible by 2 or 3 from 1 to 20 are 2, 4, 6, 8, 10, 12,
14, 16, 18, 3, 9, 15, 20 = 13

Total outcomes = 20
Possible outcomes = 13
∴ P(divisible by 2 or 3) = 13/20
ii) Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 = 8
Total Outcomes = 20
Possible outcomes = 8
∴ P(a prime number) = 8/20=2/5

Question 4) A game of chance consists of spinning an arrow on a
3 circular board, divided into / 4 8 equal parts, which comes to
rest pointing at one of the numbers 1, 2, 3, …, 8 (Figure), which
are equally likely outcomes. What is the probability that the
arrow will point at
(i) an odd number
(ii) a number greater than 3
(iii) a number less than 9.

Solution:
Total numbers = 8
(i) “Odd numbers” are 1, 3, 5, 7, i.e., 4
∴ P(an odd number) = 4/8=1/2

(ii) “nos. greater than 3” are 4, 5, 6, 7, 8, i.e., 5
∴ P(a number > 3) = 5/8

(iii) “numbers less than 9” are 1, 2, 3, …8 i.e., 8
∴ P(a number < 9) = 8/8 = 1

Question 5) A number x is selected at random from the

numbers 1, 2, 3 and 4. Another number y is selected at random
from the numbers 1, 4, 9 and 16. Find the probability that
product of x and y is less than 16.
Solution:
X can be any one of 1, 2, 3, and 4 i.e., 4 ways
Y can be any one of 1, 4, 9, and 16 i.e., 4 ways
Total no. of cases of XY = 4 × 4 = 16 ways
No. of cases, where product is less than 16 (1, 1), (1, 4), (1, 9), (2,
1), (2, 4), (3, 1), (3, 4), (4,1) i.e., 8 ways
∴ P (product x & y less then 16) = 8/16=1/2

Question 6) The probability of guessing the correct answer to a

certain question is x/12 . If the probability of guessing the wrong
answer is 3/4 , find x. If a student copies the answer, then its
probability is 2/6 . If he doesn’t copy the answer, then the
probability is 2y/3 . Find the value of y.
Solution:

P (guessing) + P (guessing wrong) = 1

Question 7)
A number x is selected at random from the numbers 1, 4, 9, 16
and another number y is selected at random from the numbers
1, 2, 3, 4. Find the probability that the value of xy is more than
16.
Solution:
x can be any one of 1, 4, 9, or 16, i.e., 4 ways
y can be any one of 1, 2, 3, or 4, i.e., 4 ways
Total number of cases of xy = 4 × 4 = 16 ways
Number of cases, where product is more than 16
(9, 2), (9, 3), (9, 4), (16, 2), (16, 3), (16, 4), i.e., 6 ways
∴ Required probability = 6/16=3/8
Question 8)
A game consists of tossing a coin 3 times and noting its outcome

each time. Hanif wins if he gets three heads or three tails, and
loses otherwise. Calculate the probability that Hanif will lose the
game.
Solution:
The possible outcomes on tossing a coin 3 times are,
S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT) = 8
Outcomes when Hanif wins = {HHH, TTT} = 2
∴ P (Hanif wins) = 2/8=1/4
∴ P (Hanif will lose) = 1 – 1/4=3/4

. Question 9) A box contains 100 red cards, 200 yellow cards

and 50 blue cards. If a card is drawn at random from the box,
then find the probability that it will be
(i) a blue card
(ii) not a yellow card
(iii) neither yellow nor a blue card.

Solution: No. of red cards = 100
No. of yellow cards = 200
No. of blue cards = 50
Total no. of cards = 100 + 200 + 50 = 350

Question 10)

i) A card is drawn from a well shuffled deck of 52 cards. Find the

probability of getting
(i) a king of red colour
(ii) a face card
(iii) the queen of diamonds.
Solution:
(i) Total cards in a deck = 52
Total no. of kings = 4
Total no. of red kings = 2

ii) All kings, queens and aces are removed from a pack of 52
cards. The remaining cards are well shuffled and then a card is
drawn from it. Find the probability that the drawn card is
(i) a black face card.
(ii) a red card.
Solution:
Total no. of cards = 52
No. of cards removed = (4 + 4 + 4) = 12

Remaining cards = 40

iii) All the red face cards are removed from a pack of 52 playing
cards. A card is drawn at random from the remaining cards, after
reshuffling them. Find the probability that the drawn card is
(i) of red colour
(ii) a queen
(iii) an ace
(iv) a face card
Solution:
Total number of cards = 52
Red face cards = 6
Remaining cards = 52 – 6 = 46

iv) Five cards, the ten, jack, queen, king and ace of diamonds, are
well shuffled with their faces downwards. One card is then
picked up at random.
(a) What is the probability that the drawn card is the queen?
(b) If the queen is drawn and put aside, and a second card is

drawn, find the probability that the second card is (i) an ace (ii) a
queen.
Solution:
(a) Total events = 5; P(queen) = 1/5
(b) Now total events = 4
(i) P (an ace) = 1/4
(ii) P (a queen) = 0/4 = 0 …[As there is no queen left
v) A card is drawn at random from a well shuffled deck of
playing cards. Find the probability that the card drawn is
(2015OD)
(i) a card of spade or an ace.
(ii) a black king.
(iii) neither a jack nor a king.
(iv) either a king or a queen.
Solution:
Total no. of outcomes = 52

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