16. A committee of 5 Students is to be chosen from 6 boys and 4 girls. Find the probability that the committee contains exactly 2 girls.
A. 11/21 B. 12/21 C. 10/21 D. 13/21
Answer: C
5 Students can be selected from 10 in 10C5 ways.
∴ n (S) = 10C5= (10!/ 5!.5!) = (10 × 9 × 8 × 7 × 6)/ (5 × 4 × 3 × 2) = 252
Let A be the event that the committee includes exactly 2 girls and 3 boys.
The two girls can be selected in 4C2 ways and the 3 boys can be selected in 6C3 ways.
∴ n(A) = 4C2 x 6C3 = 6 x 20 = 120
∴ P(A) = n(A)/n(S) = (120/252) = 10/21
17. In tossing three coins at a time, what is the probability of getting at most one head?
a) 7/8 b)1/2 c)3/8 d)1/8
Answer: (b)
Possible samples are as follows
{HHH, HTH, HHT, THH, TTH, THT, HTT, TTT}
Let A be the event of getting one head.
Let B be the event of getting no head.
Favourable outcome for
A = {TTH, THT, HTT}
Favourable outcome for
B = {TTT}
Total no. of outcomes = 8
∴ P(A) = 3/8 , P(B) = 1/8
∴ Required probability = Probability of getting one head + Probability of getting no head
= P(A) + P(B) = (3/8) + (1/8) = (4/8) = (1/2)
18. The probability that a man can hit a target is 3/4. He tries 5 times. Theprobability that he will hit the target at least three times is
A. 371/461 B. 471/502 C. 291/364 D. 459/512
Answer: (d)
Required probability is
19. The fair dice are thrown. The probability that the numbers that appear are not all distinct is
A. 4/9 B. 1/6 C. 5/9 D. 5/6
Answer: (a)
Out of the 6^3 possible outcomes, 6.5.5 outcomes will have all distinct numbers.
The probability = 1 – (6.5.4)/6^3 = 4/9
20. Two dice are thrown at a time, find the probability that the sums of the numbers on the upper faces of the dice are equal to 7.
A. 1/4 B. 1/3 C. 1/8 D. 1/6
Answer: (d)
If a die is thrown, there are 6 equally likely and mutually exclusive cases.
Since two dice are thrown, the total number of ways = 6 × 6 = 36.
If a sum of 7 is to be obtained from the numbers appearing on the two upper faces, the numbers in the two dice can be (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6,1), which are six in number.
∴ Number of favourable cases = m = 6
Total number of cases = 36
∴ The required probability = p = m / n = 6/ 36 = 1/6
