CBSE 2025, Set 1

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 that 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

Edvaya · Accepted Answer

The correct answer is Subjective. Let $E_1$ be the event of making a correct decision in committee I. Let $E_2$ be the event of making a correct decision in committee II. Since the committees are independent, events $E_1$ and $E_2$ are independent. We are given the probabilities of making a wrong decision: $P(E_1') = 0.03$ (wrong decision in committee I) $P(E_2') = 0.01$ (wrong decision in committee II) Thus, the probabilities of making a correct decision are: $P(E_1) = 1 - P(E_1') = 1 - 0.03 = 0.97$ $P(E_2) = 1 - P(E_2') = 1 - 0.01 = 0.99$ *(i) Probability of making correct decisions in both committees:* Since events are independent, $P(E_1 \cap E_2) = P(E_1) 	imes P(E_2)$. $P(E_1 \cap E_2) = 0.97 	imes 0.99$ $P(E_1 \cap E_2) = 0.9603$ *(ii) Probability of making a correct decision in only one committee:* This can happen in two mutually exclusive ways: 1. Correct decision in committee I and wrong in committee II: $P(E_1 \cap E_2')$ 2. Wrong decision in committee I and correct in committee II: $P(E_1' \cap E_2)$ Total Probability $= P(E_1 \cap E_2') + P(E_1' \cap E_2)$ Because they are independent events, we multiply the individual probabilities: $= P(E_1)P(E_2') + P(E_1')P(E_2)$ $= (0.97 	imes 0.01) + (0.03 	imes 0.99)$ $= 0.0097 + 0.0297$ $= 0.0394$

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