An urn contains two white and two black balls, A ball is drawn at random. If it is white, it is not replaced into urn otherwise it is replaced along with another ball of the same colour. The process is repeated. If ‘p’ is the probability that the third ball drawn is black. then the value of $\frac{23}{5 p}$ is

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answer is 6.

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Detailed Solution

For the first draws, following events may occur E₁ =
both balls are white
E₂ = first is white and second is black
E₃ = first is black and second is white
E₄ = both balls are black
Let represent the event that the third ball is black. Then,
$\begin{array}{l} P (E) = P (E_{1}) P (\frac{E}{E_{1}}) + \dots . + P (E_{4}) P (\frac{E}{E_{4}}) \\ = \frac{1}{6} \times \frac{3}{2} + \frac{1}{3} \times \frac{3}{4} + \frac{1}{5} \times \frac{3}{4} + \frac{3}{10} \times \frac{4}{6} = \frac{23}{30} \end{array}$