1. A coin is biased so that its chances of landing Head is 2⁄3. If the coin is flipped 3 times, the probability that the first 2 flips are heads and the 3rd flip is a tail is?
a) 4⁄27
b) 8⁄27
c) 4⁄9
d) 2⁄9
Explanation: Required probability = 2⁄3 x 2⁄3 x 1⁄3 = 4⁄27.
2. Husband and wife apply for two vacant spots in a company. If the probability of wife getting selected and husband getting selected are 3/7 and 2/3 respectively, what is the probability that neither of them will be selected?
a) 2⁄7
b) 5⁄7
c) 4⁄21
d) 17⁄21
Explanation: Let H be the event of husband getting selected
W be the event of wife getting selected
Then, the event of neither of them getting selected is = (H ∩ W)
P (H ∩ W) = P (H) x P (W)
= (1 – P (H)) x (1 – P (W))
= (1 – 2⁄3) x (1 – 3⁄7)
= 4⁄21.
3. For two events A and B, if P (B) = 0.5 and P (A ∪ B) = 0.5, then P (A|B) = ?
a) 0.5
b) 0
c) 0.25
d) 1
Explanation: We know that,
P (A│B) = P(A ∩ B)/P(B)
= P((A ∪ B)/P(B))
= (1 – P(A ∪ B)) /P(B)
= (1 – 0.5)/0.5
= 1.
4. A fair coin is tossed thrice, what is the probability of getting all 3 same outcomes?
a) 3⁄4
b) 1⁄4
c) 1⁄2
d) 1⁄6
Explanation: S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
All 3 same outcomes mean either all head or all tail
Total outcomes = 8
Favourable outcomes = {HHH, TTT} = 2
∴ required probability = 2⁄8 = 1⁄4.
5. A bag contains 5 red and 3 yellow balls. Two balls are picked at random. What is the probability that both are of the same colour?
a) \(\frac{C_2^5}{C_2^8}+\frac{C_2^3}{C_2^8}\)
b) \(\frac{C_2^5 * C_2^3}{C_2^8}\)
c) \(\frac{C_1^5 * C_1^3}{C_2^8}\)
d) 0.5
Explanation: Total no.of balls = 5R+3Y = 8
No.of ways in which 2 balls can be picked = 8C2
Probability of picking both balls as red = 5C2 /8C2
Probability of picking both balls as yellow = 3C2 /8C2
∴ required probability \(\frac{C_2^5}{C_2^8}+\frac{C_2^3}{C_2^8}\).
6. If 40% of boys opted for maths and 60% of girls opted for maths, then what is the probability that maths is chosen if half of the class’s population is girls?
a) 0.5
b) 0.6
c) 0.7
d) 0.4
Explanation: Let E be the event of electing boy or a girl and A be the event of selecting a maths student.
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2)
\(= (\frac{1}{2})(\frac{40}{100}) + (\frac{1}{2}) (\frac{60}{100}) \)
= 0.5.
7. Company A produces 10% defective products, Company B produces 20% defective products and C produces 5% defective products. If choosing a company is an equally likely event, then find the probability that the product chosen is defective.
a) 0.22
b) 0.12
c) 0.11
d) 0.21
Explanation: Let A be the event of selecting a defective item. Let Ei be the event of selecting a company. Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + P(E3) P(A|E3)
\( = (\frac{1}{3})(\frac{10}{100}) + (\frac{1}{3})(\frac{20}{100}) + (\frac{1}{3})(\frac{5}{100}) \)
\( = \frac{0.35}{3} = 0.12. \)
8. Suppose 5 men out of 100 men and 10 women out of 250 women are colour blind, then find the total probability of colour blind people. (Assume that both men and women are in equal numbers.)
a) 0.45
b) 0.045
c) 0.05
d) 0.5
Explanation: Let A be the event of selecting a colour blind person and Ei be the event of selecting a person. Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2)
\(= (\frac{1}{2})(\frac{5}{100}) + (\frac{1}{2})(\frac{10}{250}) \)
= 0.045.
9. A problem is given to 5 students P, Q, R, S, T. If the probability of solving the problem individually is 1/2, 1/3, 2/3, 1/5, 1/6 respectively, then find the probability that the problem is solved.
a) 0.47
b) 0.37
c) 0.57
d) 0.27
Explanation: Let A be the event that the problem is solved. Let Ei be the event that a student is chosen. Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + P(E3) P(A|E3) + P(E4) P(A|E4) + P(E5) P(A|E5)
\( = (\frac{1}{5})(\frac{1}{2}) + (\frac{1}{5})(\frac{1}{3}) + (\frac{1}{5})(\frac{2}{3}) + (\frac{1}{5})(\frac{1}{5}) + (\frac{1}{5})(\frac{1}{6}) \)
= 0.37.
10. The probability that the political party A does a particular work is 30% and the political party B doing the same work is 40%. Then find the probability that the work is completed if the probability of choosing the political party A is 40% and that of B is 60%.
a) 0.12
b) 0.24
c) 0.36
d) 0.48
Explanation: Let A be the event of completing the work and Ei be the event of selecting the political party. Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2)
\( = (\frac{40}{100})(\frac{30}{100}) + (\frac{60}{100})(\frac{40}{100}) \)
= 0.36.