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 $W_1$ denote the event of a wrong decision in committee I, and $W_2$ denote the event of a wrong decision in committee II. The probabilities of these events are given by: \[ P(W_1) = 0.03, \quad P(W_2) = 0.01. \] The probabilities of making a correct decision in each committee, denoted by $C_1$ and $C_2$ respectively, are: \[ P(C_1) = 1 - P(W_1) = 0.97, \quad P(C_2) = 1 - P(W_2) = 0.99. \] Calculations for the required probabilities are as follows: (i) Probability of a correct decision in both committees: Due to the independence of decisions in the two committees, this probability is the product of the individual correct decision probabilities: \[ P(C_1 \cap C_2) = P(C_1) \times P(C_2) = 0.97 \times 0.99 = 0.9603. \] (ii) Probability of a correct decision in exactly one committee: This can occur in two mutually exclusive scenarios: - Correct decision in committee I and wrong decision in committee II ($C_1 \cap W_2$). - Wrong decision in committee I and correct decision in committee II ($W_1 \cap C_2$). The probabilities for these scenarios are: \[ P(C_1 \cap W_2) = P(C_1) \times P(W_2) = 0.97 \times 0.01 = 0.0097. \] \[ P(W_1 \cap C_2) = P(W_1) \times P(C_2) = 0.03 \times 0.99 = 0.0297. \] The total probability of a correct decision in only one committee is the sum of these probabilities: \[ 0.0097 + 0.0297 = 0.0394. \] The final results 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 \( \boxed{0.0394} \).