Question

An urn contains $5$ red and $2$ green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. If the probability that the second ball is red is $\frac{a}{b}$, then $a \times b =$

Answer 1

Let $G$ represents drawing a green ball and $0$R represents drawing a red ball.
So, the probability that second drawn ball is red
$ = P(G) \cdot P(\frac{R}{G}) + P(R) P(\frac{R}{R})$
$ = \frac{2}{7} \times \frac{6}{7} + \frac{5}{7} \times \frac{4}{7} $
$= \frac{12 + 20}{49} = \frac{32}{49} = \frac{a}{b}$
$\Rightarrow a \times b = 32 \times 49 = 1568$.