Question

In a purse there are 10 coins, all 5 paise except one which is a rupee. In another purse, there are 10 coins, all 5 paise 9 coins are taken out from the former purse & put into the latter. Then 9 coins are taken out from the latter & put into the former. Then the chance that the rupee is still in the first purse is

• A. $\frac{9}{19}$
• B. $\frac{10}{19}$
• C. $\frac{4}{9}$
• D. none

Answer 1

The correct option is B $\frac{10}{19}$

Purse A has 9 - 5 paise coins and 1 - 1 rupee coin.

Purse B has 10 - 5 paise coins.

When all 5 paise 9 coins are taken out from the former purse & put into the latter, there are 2 cases,

case (i): from A : 9 - 5 paise coins into B $⟹$ purse A has 1 - 1 rupee coin and purse B has 19 - 5 paise coins. The probability of choosing 9 - paise coin is $\frac{1}{10}$

case (ii): from A: 8 - 5 paise coins and 1 - 1 rupee coin $⟹$ purse A has 1 - 5 paise coin and purse B has 18 - 5 paise coins and 1 - 1 rupee coin. The probability here becomes $\frac{9}{10}$

Then 9 coins are taken out from the latter & put into the former, so possible cases are:

case (i): from B: 9 - 5 paise coins into A $⟹$ purse A has 1 - 1 rupee coin and 9 - 5 paise coins. No change. Here the probability is $\frac{1}{10}$

case (ii): (a) from B: 8 - 5 paise coins and 1 - 1 rupee coin $⟹$ purse A has 9 - 5 paise coins and 1 - 1 rupee coin. No change. Here the probability is $\frac{{}^{18}{C}_8}{{}^{19}{C}_9}$

case (ii): (b) from B: 9 - 5 paise coins into A $⟹$ purse A has 10 - 5 paise coins. Here probability is $\frac{{}^{18}{C}_9}{{}^{19}{C}_9}$

So P(chance that the rupee is still in the first purse ) = $\frac{1}{10}+\frac{9}{10}\times \frac{{}^{18}{C}_8}{{}^{19}{C}_9}$

$⟹=\frac{1}{10}+\frac{9}{10}\times \frac{18!}{(18-8)!8!}\times \frac{(19-9)!9!}{19!}⟹=\frac{1}{10}+\frac{9}{10}\times \frac{9}{19}⟹=\frac{1}{10}+\frac{81}{190}⟹=\frac{19+81}{190}⟹\frac{100}{190}=\frac{10}{19}$