A box contains 5 black and 4 white balls. A ball is drawn at random and its colour is noted. The ball is then put back in the box along with two additional balls of its opposite colour. If a ball is drawn again from the box, then the probability that the ball drawn now is black, is

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$\frac{5}{11}$

$\frac{53}{99}$

$\frac{48}{99}$

$\frac{7}{11}$

answer is B.

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Detailed Solution

Let $E_{1}$ : black ball is drawn in the first draw $E_{2}$ : white ball is drawn in the first draw $A$ : a black ball is drawn in the second draw

$\begin{array}{l} P (E_{1}) = \frac{5}{9}, P (E_{2}) = \frac{4}{9} \\ P (A ∣ E_{1}) = \frac{5}{9 + 2}, P (A ∣ E_{2}) = \frac{5 + 2}{9 + 2} \end{array}$

By the total probability rule
$\begin{array}{l} P (A) = P (E_{1}) P (A ∣ E_{1}) + P (E_{2}) P (A ∣ E_{2}) \\ = (\frac{5}{9}) (\frac{5}{11}) + (\frac{4}{9}) (\frac{7}{11}) = \frac{53}{99} \end{array}$