Question

A person is Head of two independent selection committees I and II. If the probability of making a wrong selection in committee I is 0.03 and in committee II is 0.01, then find the probability that the person makes the correct decision of selection: (i) in both committees (ii) in only one committee

Hint:

To solve probability problems involving independent events, remember to multiply the probabilities for the combined event. For events happening in one committee, you can compute the probability for each case and sum them up.

Answer 1

Let the event of making a wrong decision in committee I be denoted as $W_1$ and the event of making a wrong decision in committee II be denoted as $W_2$. The probabilities of making a wrong decision are given as: \[ P(W_1) = 0.03, \quad P(W_2) = 0.01. \] Since the person is making the correct decision in each committee if they do not make a wrong decision, the probability of making the correct decision in each committee is: \[ P(C_1) = 1 - P(W_1) = 1 - 0.03 = 0.97, \quad P(C_2) = 1 - P(W_2) = 1 - 0.01 = 0.99. \] Now, let's compute the required probabilities. (i) Probability that the person makes the correct decision in both committees: Since the decisions in the two committees are independent, the probability of making the correct decision in both committees is the product of the probabilities of making the correct decision in each committee: \[ P(C_1 \cap C_2) = P(C_1) \times P(C_2) = 0.97 \times 0.99 = 0.9603. \] (ii) Probability that the person makes the correct decision in only one committee: The person can make the correct decision in only one committee in two ways: - The person makes the correct decision in committee I but the wrong decision in committee II. - The person makes the wrong decision in committee I but the correct decision in committee II. The probability of making the correct decision in committee I but the wrong decision in committee II is: \[ P(C_1 \cap W_2) = P(C_1) \times P(W_2) = 0.97 \times 0.01 = 0.0097. \] The probability of making the wrong decision in committee I but the correct decision in committee II is: \[ P(W_1 \cap C_2) = P(W_1) \times P(C_2) = 0.03 \times 0.99 = 0.0297. \] Thus, the probability of making the correct decision in only one committee is: \[ P(C_1 \cap W_2) + P(W_1 \cap C_2) = 0.0097 + 0.0297 = 0.0394. \] Thus, the answers are: (i) The probability of making the correct decision in both committees is \( \boxed{0.9603} \). (ii) The probability of making the correct decision in only one committee is \( \boxed{0.0394} \).