This set of Probability and Statistics Multiple Choice Questions & Answers (MCQs) focuses on “Theorem of Total Probability”.
1. 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
View Answer
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.
2. 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
View Answer
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)
\( = (\frac{1}{3})(\frac{10}{100}) + (\frac{1}{3})(\frac{20}{100}) + (\frac{1}{3})(\frac{5}{100}) \)
\( = \frac{0.35}{3} = 0.12. \)
3. 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
View Answer
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.
4. 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) 0.93
c) 0.57
d) 0.27
View Answer
Explanation: The problem will be solved if at least one student solves it. We can calculate the probability of the problem not being solved and then subtract that from 1 to get the probability of it being solved.
Probability (Problem is solved) = 1 – Probability (No one solved the problem)
= 1 – ((1/2)*(2/3)*(1/3)*(4/5)*(5/6))
= 25/27 = 0.93
5. 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
View Answer
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.
6. Total probability theorem is used in Baye’s theorem.
a) True
b) False
View Answer
Explanation: Total probability theorem is used in Baye’s theorem. Baye’s theorem is given by
\(P(E_{i}|A) = \frac{P(Ei)P(A∨Ei)}{∑P (Ei)P(A∨Ei)}. \)
7. Theorem of total probability is given by P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) +…..(n terms).
a) True
b) False
View Answer
Explanation: The total probability P(A) is given by the sum of the product of the probability of a particular event and the probability of A given the particular event. Hence the formula is true.
8. In badminton practice session, the probability that the player A serves properly is 0.8 and that he player B serves properly is 0.9. If there are only two players, then find the probability that it is serves properly.
a) 0.75
b) 0.85
c) 0.95
d) 0.55
View Answer
Explanation: Let A be the event that the service is done properly and Ei be the event that a player is selected. Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2)
\( = (\frac{1}{2})(\frac{8}{10}) + (\frac{1}{2})(\frac{9}{10}) \)
= 0.85.
9. The probability that person A completes all the tasks assigned is 50% and that of person B is 20%. Find the probability that all the tasks are completed.
a) 0.15
b) 0.25
c) 0.35
d) 0.45
View Answer
Explanation: Let A be the event that all tasks are completed and Ei is the event that a person is selected.
Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2)
\( =(\frac{1}{2})(\frac{50}{100}) + (\frac{1}{2})(\frac{20}{100}) \)
= 0.35
10. Let there be two newly launched phones A and B. The probability that phone A has good battery life is 0.7 and the probability that phone B has good battery life is 0.8. Then find the probability that a phone has a good battery life.
a) 0.65
b) 0.75
c) 0.85
d) 0.45
View Answer
Explanation: Let A be the event thtat a phome has a good battery life and Ei be the event that a phone is selected. Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2)
\( = (\frac{1}{2})(0.7) + (\frac{1}{2})(0.8) \)
= 0.75.
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