Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected ball is black, given that the second selected ball is also black, is $\frac{m}{n}$ where gcd(m, n) = 1, then m + n is equal to :

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Detailed Solution

$n (E) = \frac{6}{10} \times \frac{5}{9}$

$n (S) = \frac{4}{10} \times \frac{6}{9} + \frac{6}{10} \times \frac{5}{9}$

$\begin{aligned} P (E) = \frac{\frac{6}{10} \times \frac{5}{9}}{\frac{4}{10} \times \frac{6}{9} + \frac{6}{10} \times \frac{5}{9}} = \frac{5}{9} \end{aligned}$

$\begin{array}{r} m = 5, n = 9 \end{array}$

$m + n = 14$

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