Question

In a purse there are 10 coins, all five $5$ paisa coins except one which is a rupee; in another there are 10 coins all five $5$ paisa .Nine coins are takes from the former and put into the latter and then nine coins are taken from the latter and put into the former, the chance that the rupee is still in the first purse is $1-\frac{k}{19}$. Find the value of $k$.

Answer 1

To find : Value of $k$

There are two cases

1.$Re1$ coin is not transferred in first attempt

2.$Re1$ coin is transferred in first attempt and retransferred in second attempt

So probability is $P_1+P_2$

$=\frac{1}{10}+$(probability that in first attempt Re 1 is transferred)$\times$(probability it is retransferred)

$=\frac{1}{10}+\frac{9}{10}\times {\frac{{}^1{C}_1{\times }^{18}{C}_8}{{}^{19}{C}_9}}_{}$

Out of 18 5 paisa coins 8 should be selected and out of 19 only 9 should be transferred.

$=\frac{1}{10}+\frac{9}{10}\times \frac{18!\times 9!}{8!\times 19!}=\frac{1}{10}+\frac{9}{10}\times \frac{9}{19}$

$=\frac{19+81}{190}=\frac{10}{19}=1-\frac{9}{19}$

So $k=9$