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 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 $1 - \dfrac{k}{{19}}$ .Find the value of $k$ .

Answer 1

Hint : Here, the word problem is converted into mathematical expression and we have to find out the value of $k$ by using a probability method. Probability is the extent to which an event is likely to occur, measured by the ratio of the favourable cases to the whole number of cases possible.

Complete step by step solution:
In the given problem,
There are $10$ coins in purse $A$ , in which nine $5$ paisa coins and one $1$ rupee coin
There are $10$ coins in purse $B$ , in which ten $5$ paisa coins
When nine coins are taken from the former and put into the latter and the nine coins are taken from the latter and put into the former, then there are two cases
Case 1: one rupee coin is not transferred in the first attempt
Case 2: one rupee coin is transferred in the first attempt and retransferred in the second attempt
Therefore, the probability is ${P_A} + {P_B}$
$= \dfrac{1}{{10}} +$ [Probability of one rupee coin is transferred $\times$ probability of one rupee coin is retransferred]
$= \dfrac{1}{{10}} + [\dfrac{9}{{10}} \times \dfrac{{^1{C_1}{ \times ^{18}}{C_8}}}{{^{19}{C_9}}}]$
Here,
$^{18}{C_8}$ defines that among $18$ five paisa coins $8$ coins should be transferred
$^{19}{C_9}$ defines that among $19$ five paisa coins $9$ coins should be transferred
On comparing $^{18}{C_8}$ and $^{19}{C_9}$ with the formula of $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$ , we can get
$= \dfrac{1}{{10}} + \dfrac{9}{{10}}[\dfrac{{\dfrac{{18!}}{{(18 - 8)!8!}}}}{{\dfrac{{19!}}{{(19 - 9)!9!}}}}]$
$= \dfrac{1}{{10}} + \dfrac{9}{{10}}[\dfrac{{18!}}{{10!8!}} \times \dfrac{{10!9!}}{{19!}}]$
By simplifying the factors, we can get
$= \dfrac{1}{{10}} + \dfrac{9}{{10}}[\dfrac{9}{{19}}]$
$= \dfrac{1}{{10}} + \dfrac{{81}}{{190}}$
Taking LCM, we get
$= \dfrac{{19 + 81}}{{190}}$
$= \dfrac{{100}}{{190}}$
Finally, we get
$= \dfrac{{10}}{{19}}$
If the rupee is still in the first purse, there is a chance of $1 - \dfrac{k}{{19}}$
Therefore, $\dfrac{{10}}{{19}} = 1 - \dfrac{9}{{19}}$
On comparing $1 - \dfrac{9}{{19}}$ with the given form of $1 - \dfrac{k}{{19}}$ , we can get the value of $k$
Hence, $k = 9$
So, the correct answer is “ $k = 9$ ”.

Note : In this coin problem, we have converted the word problem into mathematical expression and we have used the probability method and the combination formula of $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$ to get the value of $k$ . Probability is used to mean the chance that the particular event will occur.