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

Answer: 1568

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