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Let $P(W_1)$ be the probability of making a wrong selection in committee I, and $P(W_2)$ be the probability of making a wrong selection in committee II. We are given: $P(W_1) = 0.03$ $P(W_2) = 0.01$
Let $P(C_1)$ be the probability of making a correct selection in committee I, and $P(C_2)$ be the probability of making a correct selection in committee II. Then: $P(C_1) = 1 - P(W_1) = 1 - 0.03 = 0.97$ $P(C_2) = 1 - P(W_2) = 1 - 0.01 = 0.99$
Since the committees are independent, the probability of making the correct selection in both committees is: $P(C_1 \cap C_2) = P(C_1) \times P(C_2) = 0.97 \times 0.99 = 0.9603$
The probability of making the correct selection in only one committee can be calculated as the sum of the probabilities of making a correct selection in committee I and a wrong selection in committee II, and making a wrong selection in committee I and a correct selection in committee II. $P(\text{Only one correct}) = P(C_1 \cap W_2) + P(W_1 \cap C_2) = P(C_1)P(W_2) + P(W_1)P(C_2)$ $P(\text{Only one correct}) = (0.97)(0.01) + (0.03)(0.99) = 0.0097 + 0.0297 = 0.0394$
Final Answer: (i) 0.9603 (ii) 0.0394