# ICSE Solutions for Chapter 25 Probability Class 10 Mathematics

Question 1: A bag contains 3 red balls, 4 blue balls and 1 yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it; find the probability that the ball is:

(i) yellow
(ii) red
(iii) not yellow
(iv) neither yellow nor red

Solution 1: Total number of balls in the bag  = 3 + 4 + 1 = 8 balls
Number of possible outcomes = 8 = n(S)
(i) Event of drawing a yellow ball ={Y}
n(E) = 1
Probability of drawing a yellow ball = n(E)/n(S) = 1/8

(ii) Event of drawing a red ball = {R, R, R]
n(E) = 3
Probability of drawing a red ball = n(E)/n(S) = 3/8

(iii) Probability of not drawing a yellow ball = 1 - Probability of drawing a yellow ball
Probability of not drawing a yellow ball  = 1 – 1/8
= (8 – 1)8 = 7/8

(iv) Neither yellow ball nor red ball means a blue ball
Event of not drawing a yellow or red ball = E = 4
n(E) = 4
Probability of not drawing a yellow or red ball = n(E)/n(S) = 4/8 = 1/2

Question 2: A dice is thrown once. What is the probability of getting a number:
(i) greater than 2?
(ii) less than or equal to 2?

Solution 2: Number of possible outcomes when dice is thrown  = (1, 2, 3, 4, 5, 6)
n(S) = 6
(i) Event of getting a number greater than 2 = E  = (3, 4, 5, 6)
n(E) = 4
Probability of getting a number greater than 2 = n(E)/n(S) = 4/6 = 2/3

(ii) Event of getting a number less than or equal to 2 = E = (1, 2)
n(E) = 2
Probability of getting a number less than or equal to 2 =
n(E)/n(S) = 2/6 = 1/3

Question 3: From 25 identical cards, numbered 1, 2, 3, 4, 5.....24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 3
(ii) 5
(iii) 3 and 5
(iv) 3 or 5

Solution 3: There are 25 cards from which one card is drawn.
Total number of elementary events = n(S) = 25

(i) From numbers 1 to 25, there are 8 numbers which are multiple of 3 i.e. (3, 6, 9, 12, 15, 18, 21, 24) Favorable number of events = n(E) = 8
Probability of selecting a card with a multiple of 3 = n(E)/n(S) = 8/25

(ii) From numbers 1 to 25, there are 5 numbers which are multiple of 3 and 5 i.e. (5, 10, 15, 20, 25) Favorable number of events = n(E) = 5
Probability of selecting a card with a multiple of 5 = n(E)/n(S) = 5/25 = 1/5

(iii) From numbers 1 to 25, there is only one number which is multiple of 3 and 5 i.e. (15) Favorable number of events = n(E) = 1
Probability of selecting a card with a multiple of 3 and 5 =
n(E)/n(S) = 1/25

(iv) From numbers 1 to 25, there are 12 numbers which are multiple of 3 or 5 i.e. (3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25) Favorable number of events = n(E) = 12
Probability of selecting a card with a multiple of 3 or 5 = n(E)/n(S) =  25

Question 4: A die is thrown once. Find the probability of getting a number:
(i) less than 3
(ii) greater than or equal to 4
(iii) less than 8
(iv) greater than 6

Solution 4: In throwing a dice, total possible outcomes = {1 ,2, 3, 4, 5, 6)
n(S) = 6

(i) On a dice, numbers less than 3 = {1, 2}
n(E) = 2
Probability of getting a number less than 3 = n(E)/n(S) = 2/6 = 1/3

(ii) On a dice, numbers greater than or equal to 4 =  {4, 5, 6}
n(E) = 3
Probability of getting a number greater than or equal to 4 = n(E)/n(S) = 3/6 = 1/2

(iii) On a dice, numbers less than 8 = {1, 2, 3, 4, 5, 6}
n(E) = 6
Probability of getting a number less than 8 = n(E)/n(S) = 6/6 = 1

(iv) On a dice, numbers greater than 6 = 0
n(E) = 0
Probability of getting a number greater than 6 = n(E)/n(S)  = 0/6 = 0

Question 5: From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will:
(i) be a black card
(ii) not be a red card.
(iii) be a red card.
(iv) be a face card
(v) be a face card of red colour

Solution 5: Total number of cards = 52
Total number of outcomes = P(s) = 52
There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and diamond are black. So, there are 26 red cards and 26 black cards.

(i) Number of black cards in a deck = 26
P(E) = favourable outcomes for the event of drawing a black card = 26
Probability of drawing a black card = P(E)/P(S) = 26/52 = ½

(ii) Number of red cards in a deck = 26
Therefore, number of non-red cards = 52 – 26 = 26
P(E) = favourable outcomes for the event of not drawing a red card = 26
Probability of not drawing a red card = P(E)/P(S) = 26/52 = ½

(iii) Number of red cards in a deck = 26
P(E) = favourable outcomes for the event of drawing a red card = 26
Probability of drawing a red card = P(E)/P(S) = 26/52 = 1/2

(iv) There are 52 cards in a deck of cards, and 12 of these cards are face cards (4 kings, 4 queens, and 4 jacks).
P(E) = 12
Probability of drawing a face card = P(E)/P(S) = 12/52 = 3/13

(v) There are 26 red cards in a deck, and 6 of these cards are face cards (2 kings, 2 queens, and 2 jacks).
P(E) = 6
Probability of drawing a red face card = P(E)/P(S) = 6/52 = 3/26

Question 6:
(i) If A and B are two complementary events then what is the relation between P(A) and P(B)?
(ii) If the probability of happening an event A is 0.46. What will be the probability of not happening of the event A?

Solution 6:
(i) Two complimentary events, taken together, include all the outcomes for an experiment and the sum of the probabilities of all outcomes is 1.
P(A) + P(B) = 1

(ii) P(A) = 0.46
Let P(B) be the probability of not happening of event A
We know,
P(A) + P(B) = 1
P(B) = 1 - P(A)
P(B) = 1 - 0.46
P(B) = 0.54
Hence the probability of not happening of event A is 0.54

Question 7: In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of:
(i) winning of Geeta
(ii) not winning of Ritu

Solution 7:
(i) Winning of Geeta is a complimentary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(winning of Geeta) = 1
P(winning of Geeta) = 1 - P(winning of Ritu)
P(winning of Geeta) = 1 - 0.73
P(winning of Geeta) = 0.27

(ii) Not winning of Ritu is a complementary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(not winning of Ritu) = 1
P(not winning of Ritu) = 1 - P(winning of Ritu)
P(not winning of Ritu) = 1 - 0.73
P(not winning of Ritu) = 0.27

Question 8: Two dice are rolled together. Find the probability of getting
(i) a total of at least 10.
(ii) a multiple of 2 on one die and an odd number on the other die.

Solution 8: In throwing a dice, total possible outcomes = (1, 2, 3, 4, 5, 6)
n(S) = 6
For two dice, n(S) = 6 x 6 = 36

(i) E = event of getting a total of at least 10 = [(4, 6), (5, 5), (5, 6), (6, 4),(6, 5), (6, 6)]
n(E) = 6
Probability of getting a total of at least 10 = n(E)/n(S) = 6/36 = 1/6

(ii) E = event of getting a multiple of 2 on one die and an odd number on the other = {(2, 1), (2, 3), (2, 5),(4, 1),(4, 3), (4, 5), (6, 1), (6, 3), (6, 5), (1, 2), (3, 2), (5, 2), (1, 4),(3, 4), (5, 4),(1, 6), (3, 6), (5,6)]
n(E) = 18
Probability of getting a multiple of 2 on one die and an odd number on the other = n(E)/n(S) = 18/36 = 1/2

Question 9: In a musical chairs game, a person has been advised to stop playing the music at any time within 40 seconds after its start. What is the probability that the music will stop within the first 15 seconds?

Solution 9:
Total result = 0 sec to 40 sec
Total possible outcomes = 40
n(S) = 40
Favorable results = 0 sec to 15 sec
Favorable outcomes = 15
n(E) = 15
Probability that the music will stop in first 15 sec = n(E)/n(S) = 15/40 = 3/8

Question 10: In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that:
(i) it is acceptable to a trader who accepts only a good shirt?
(ii) it is acceptable to a trader who rejects only a shirt with major defects?

Solution 10:
Total number of shirts = 50
Total number of elementary events = 50 = n(S)

(i) Since trader accepts only good shirts and number of good shirts  = 44
Event of accepting good shirts = 44 = n(E)
Probability of accepting a good shirt = n(E)/n(S) = 44/50 = 22/25

(ii) Since trader rejects shirts with major defects only and number of shirts with major defects = 2
Event of accepting shirts = 50 - 2 = 48  = n(E)
Probability of accepting shirts = n(E)/n(S) =  48/50 =  24/25