Class 10 Maths Chapter 14 Case Based Questions - Probability

Study Case - 1

Sanjeev is very found of collecting balls of different colours. He has a total of 25 balls in his basket out of which five balls are red in colour and eight are white. Out of the remaining balls, some are green in colour and the rest are pink:      
Q1: If the probability of drawing a pink ball is twice the probability of drawing a green ball, then the number of pink balls is:
(a) 4
(b) 8
(c) 10
(d) 12
Ans: (b)
Explanation: As the total number of balls is 25 and number of red balls + white balls is 12.
∴ Total number of green balls + pink balls = 25 − 13 = 12
Let the number of pink balls be x.
Then the number of green balls = 12 − x.
We know. Probability of an event E is given by

Therefore, number of pink balls = 8.

Q2: The probability of drawing a ball of colour other than green colour is:
(a) 0
(b) 4/25
(c) 21/25
(d) 17/25
Ans: (c)
Explanation: From part (A), number of green balls = 4.
∴ Number of balls of colour other than
= 25 − 4 = 21.
∴ Probability of drawing a ball of colour other than green colour = 21/25.

Q3: The probability of drawing either a green or white ball is:
(a) 0
(b) 12/25
(c) 13/25
(d) 17/25
Ans: (b)
Explanation: The number of green balls = 4 and number of white balls =8.
Therefore, total number of green balls + white balls = 4 + 8 = 12..
∴ Probability of drawing either a or a white ball = 12/25.

Q4: The probability of drawing neither a pink nor a white ball is:
(a) 9/25
(b) 12/25
(c) 16/25
(d) 17/25
Ans: (a)
Explanation: The number of pink balls = 8
and number of white balls = 8.
∴ Total number of pink balls + white balls = 8 + 8 = 16.
∴ Probability of drawing either a pink ball or a white ball is 16/25.

∴ Probability of drawing neither a pink ball nor white ball

Q5: The probability of not drawing a red ball is:
(a) 1/5
(b) 2/5
(c) 3/5
(d) 4/5
Ans: (d)
Explanation: Number of red balls = 5.
∴ Probability of drawing a red ball = 5/25 = 1/5.

Study Case - 2

Three persons toss 3 coins simultaneously and note the outcomes. Then, they ask few questions to one another. Help them in finding the answer of the following questions:

Q1: The probability of getting almost one tail is:
(a) 0
(b) 1
(c) 1/2
(d) 1/4
Ans: (c)
Explanation: Sample space
(S) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} ⇒ n(S) = 8
Let A be the event of getting at most one tail.            
∴ A = {HHH, HHT, HTH, THH}
⇒ n(A) = 4
∴ Required probability = 4/8 = 1/2

Q2: The probability of getting exactly 1 head is:
(a) 1/2
(b) 1/4
(c) 1/8
(d) 3/8
Ans: (d)
Explanation: Let B be the event of getting exactly 1 head.
∴B = {HTT, THT, TTH}
⇒ n(B) = 3
∴ Required probability = 3/8

Q3: The probability of getting exactly 3 tails is:
(a) 0
(b) 1
(c) 1/4
(d) 1/8
Ans: (d)
Explanation: Let C be the event of getting exactly 3 tails.                
∴ C = {TTT} ⇒ n(C) = 1
∴  Required probability = 1/8

Q4: The probability of getting at most 3 heads is:  
(a) 0
(b) 1
(c) 1/2
(d) 1/8
Ans: (b)
Explanation: Let D be the event of getting atmost 3 heads.            
∴ D = {HHH, HHT, HTH, HTT, THH, THT, TTT} ⇒ n(D) = 8
∴  Required probability = 8/8 = 1.

Q5: The probability of getting at least two heads is:
(a) 0
(b) 1
(c) 1/2
(d) 1/4
Ans: (c)
Explanation: Let E be the event of getting at least two heads.
∴E = {HHT, HTH, THH, HHH}
⇒ n(E) = 4
∴ Required probability = n(E)/n(S) = 4/8 = 1/2.