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. $\dfrac{9}{19}$
B. $\dfrac{10}{19}$
C. $\dfrac{4}{9}$
D. none

Answer 1

Hint : Find the probability of selecting only the 5 paise coin from the first purse and then find the probability of selecting a one-rupee coin from the first purse and then probability of selecting the same one-rupee coin from the second purse to transfer it back to the first purse. Add both the probabilities to get the final probability.

Complete step-by-step answer :
In the question, it is given that the First purse has 10 coins, all 5 paise except one which is a rupee. Also, in the second purse, there are 10 coins, all 5 paise. Now, 9 coins are taken out from the first purse & put into the second purse. Then 9 coins are taken out from the second purse and put into the first purse. Then we have to find the probability or the chance that the rupee is still in the first purse. So here we will first find the probability of selecting only the 5 paise coins from the first purse, so that there is no effect in the one-rupee coin in the first purse. So the probability of selecting nine 5 paise coins from the first purse is $\dfrac{1}{10}$, as there is only one 1 rupee coin from 10 coins.
In the second case, when one of the coin is one-rupee coin being selected from the first purse, then we will find the probability of selecting the one-rupee coin from the first purse and then selecting the same one-rupees coin from the second purse so that the one-rupee coin is transfer back to the first purse, is to be found. So now selecting the 1-rupee coin from the first purse is $\dfrac{9}{10}$. So now the second purse will have eighteen 5 paise coins and 1 one-rupee coin. So now, the probability of selecting the one-rupee coin from the second purse is $\dfrac{{}^{18}{{C}_{8}}}{{}^{19}{{C}_{9}}}$.
Now, the probability of the second case where we are selecting the one-rupee coin from first purse and then selecting same one-rupee coin from the second purse will be $\dfrac{9}{10}\times \dfrac{{}^{18}{{C}_{8}}}{{}^{19}{{C}_{9}}}$.
So the final probability will be the sum of probabilities in the first case plus the probability of the second case, which is as follows:

& \Rightarrow \dfrac{1}{10}+\dfrac{9}{10}\times \dfrac{{}^{18}{{C}_{8}}}{{}^{19}{{C}_{9}}} \\\ & \Rightarrow \dfrac{1}{10}+\dfrac{9}{10}\times \dfrac{18!}{10!\times 8!}\times \dfrac{9!\times 10!}{19!} \\\ & \Rightarrow \dfrac{1}{10}+\dfrac{9}{10}\times \dfrac{9}{19} \\\ & \Rightarrow \dfrac{1}{10}+\dfrac{81}{190} \\\ & \Rightarrow \dfrac{100}{190} \\\ & \Rightarrow \dfrac{10}{19} \\\ \end{aligned}$$ So, the final probability is $$\dfrac{10}{19}$$. Hence the correct answer is option B. **Note** : When we are finding the probability then we can use the section method, because we know that the probability is the number of favourable cases divided by the total number of cases. Here we have to be careful that we are not using the permutation because this will give the ways of arranging coins and not selecting coins.