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(C) Given,$P(A) = 0.7$. Probability that exactly one of them is selected is $P(A \cap \overline{B}) + P(\overline{A} \cap B) = 0.6$. Since $A$ and $B$

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$0.25$

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(C) Given,$P(A) = 0.7$. Probability that exactly one of them is selected is $P(A \cap \overline{B}) + P(\overline{A} \cap B) = 0.6$. Since $A$ and $B$ are independent events,$P(A \cap B) = P(A) \cdot P(B)$. The probability that exactly one is selected is given by $P(A) + P(B) - 2P(A \cap B) = 0.6$. Substituting the values: $0.7 + P(B) - 2(0.7)P(B) = 0.6$. $0.7 + P(B) - 1.4P(B) = 0.6$. $-0.4P(B) = 0.6 - 0.7$. $-0.4P(B) = -0.1$. $P(B) = \frac{0.1}{0.4} = 0.25$. Thus,the probability that $B$ is selected is $0.25$.

$A$ and $B$ are two candidates seeking admission in a college. The probability that $A$ is selected is $0.7$ and the probability that exactly one of them is selected is $0.6$. Find the probability that $B$ is selected.

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