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 more probable of two events is

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$\frac{4}{7}$

$\frac{6}{7}$

$\frac{3}{7}$

$\frac{5}{7}$

answer is A.

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Detailed Solution

A, B are independent events

Let $P (A) = x P (B) = y$

$\begin{array}{rcl} P (A \cap B) + P (A \cap B) & = & \frac{26}{49} \\ x (1 - y) + y (1 - x) & = & \frac{26}{49} \\ x + y - 2 xy & = & \frac{26}{49} - (1) \\ P (A \cap B) & = & \frac{15}{49} \Rightarrow (1 - x) (1 - y) = \frac{15}{49} \\ 1 - x - y + xy & = & \frac{15}{49} - (2) \\ (1) + (2) & \Rightarrow & 1 - xy = \frac{41}{49} \Rightarrow xy = \frac{8}{49} \\ x & = & \frac{4}{7} y = \frac{2}{7} \end{array}$