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(A) Let $P(A) = a$ and $P(B) = b$ be the probabilities of two independent events $A$ and $B$.Since $A$ and $B$ are independent,$P(A \cap B) = ab$.The

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(A) Let $P(A) = a$ and $P(B) = b$ be the probabilities of two independent events $A$ and $B$.Since $A$ and $B$ are independent,$P(A \cap B) = ab$.The probability that exactly one of them occurs is $P(A \cap B^c) + P(A^c \cap B) = a(1-b) + b(1-a) = a + b - 2ab = \frac{26}{49}$ ... $(i)$.The probability that none of them occurs is $P(A^c \cap B^c) = (1-a)(1-b) = 1 - (a+b) + ab = \frac{15}{49}$.Thus,$a+b - ab = 1 - \frac{15}{49} = \frac{34}{49}$ ... $(ii)$.Subtracting $(i)$ from $(ii)$,we get $ab = \frac{34}{49} - \frac{26}{49} = \frac{8}{49}$.Substituting $ab$ in $(ii)$,$a+b = \frac{34}{49} + \frac{8}{49} = \frac{42}{49} = \frac{6}{7}$.Now,$a$ and $b$ are roots of the quadratic equation $x^2 - (a+b)x + ab = 0$,which is $x^2 - \frac{6}{7}x + \frac{8}{49} = 0$.Multiplying by $49$,$49x^2 - 42x + 8 = 0$.$(7x-2)(7x-4) = 0$,so $x = \frac{2}{7}$ or $x = \frac{4}{7}$.The probabilities are $\frac{2}{7}$ and $\frac{4}{7}$.The more probable event has probability $\frac{4}{7}$.

Given two independent events,if the probability that exactly one of them occurs is $\frac{26}{49}$ and the probability that none of them occurs is $\frac{15}{49}$,then the probability of the more probable of the two events is (in $/7$)

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