OP Malhotra Class 11 Maths Solutions Chapter 22 Probability Chapter Test

Practicing S Chand ISC Maths Class 11 Solutions Chapter 22 Probability Chapter Test is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Chapter Test

Question 1.
In rolling two fair dice, what is the probability of obtaining a sum greater than 3 but not exceeding 6 ?
Solution:
In rolling two dice, total no. of outcomes = n (S) = 6² = 36
Let A : event of obtaining a sum greater than 3 but not exceeding 6 = {(1,3), (2, 2), (3, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1,5), (2, 4), (4, 2), (3,3), (5,1)}
∴ n( A) =12
Thus reqd/ probability = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{12}{36}\) = \(\frac{1}{3}\)

Question 2.
Find the probability of obtaining a sum 8 in a single throw of two dice.
Solution:
In a single throw of two dice
Then total no. of outcomes = n (S) = 6² = 36
Let A : obtaining a sum 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
∴ n (A) = 5
Thus required probability
= P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{5}{36}\)

Question 3.
A bag contains 4 red, 6 white and 5 black balls. 2 balls are drawn at random. Find the probability of getting one red and one Whitehall.
Solution:
Given no. of red balls = 4, no. of white balls = 6 and no. of black balls = 5
Total no. of balls = 4 + 6 + 5 = 15
∴ n (S) = Total no. of ways of drawing 2 balls out of 15 = ¹⁵C₂
Let A : getting one red and one white ball
∴ n (A) = Total no. of ways of drawing one red ball out of 4 and 1 white ball out of
6 = ⁴C₁ × ⁶C₁
Thus required probability
= P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{{ }^4 C_1 \times{ }^6 C_1}{{ }^{15} C_2}\) = \(\frac{\frac{4 \times 6}{15 \times 14}}{2}\) = \(\frac{24}{105}\) = \(\frac{8}{35}\)

Question 4.
Out of 26 cards numbered from 1 to 26, one card is chosen. Find the probability that it is not divisible by 4.
Solution:
Total no. of outcomes = n (S) = 26
Let A : number on card is divisible by 4 = {4, 8, 12, 16, 20, 24}
∴ n (A) = 6
Thus, P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{6}{26}\) = \(\frac{3}{13}\)
∴ required probability that chosen card bear a number which is not divisible by 4
= \(P(\bar{A})\) = 1 – P (A) = 1 – \(\frac{3}{13}\) = \(\frac{10}{13}\)

Question 5.
If A and B are mutually exclusive events with \(\frac{1}{2}\) = P (B) and A ∪ B = S, the sample space. Find P (A).
Solution:
Given A and B are mutually exclusive events
∴ A ∩ B = Φ ⇒ P (A ∩ B) = 0
Given \(\frac{1}{2}\) = P (B)
Also S = A ∪ B ⇒ P(S) = P (A ∪ B) ⇒ P (A ∪ B) = 1 = P (A) + P (B) – P (A ∩ B)
⇒ 1 = P (A) + \(\frac{1}{2}\) – 0 ⇒ P (A) = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)

Question 6.
A and B are two events, such that
P (A) = 0.42, P (B) = 0.48, P (A and B) = 0.16 Determine (i) P (not A), (ii) P (not B), (iii) P (A or B)
Solution:
Given P (A) = 0.42 ; P (B) = 0.48 ; P (A and B) = P (A ∩ B) = 0.16
(i) P(not A) = \(P(\bar{A})\) = 1 – P (A) = 1 – 0.42 = 0.58
(ii) P (not B) = \(P(\bar{B})\) = l – P (B) = 1 – 0.48 = 0.52
(iii) P(A or B) = P (A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.42 + 0.48 – 0.16 = 0.9 – 0.16 = 0.74