Question

If \[mn\] coins have been distributed into $na$ purses, $n$ into each, find
(1) the chance that two specified coins will be found in the same purse;
(2) what the chance becomes when $r$ purses have been examined and found not to contain either of the specified coins.

Answer 1

Hint:
Here we are asked to find the chance in two cases. The probability or chance of an event can be found by dividing the favourable number of outcomes by total number of outcomes. So we need to find the favourable and total separately in each case.

Useful formula:
Probability of an event is calculated by dividing the number of favourable outcomes by total number of outcomes.
If the chance of an event happening is $p$, then the chance of that event not happening is $1 - p$.

Complete step by step solution:
Given that \[mn\] coins have been distributed into $na$ purses, $n$ into each.
Therefore, we have $\dfrac{{mn}}{{na}} = \dfrac{m}{a} = n$
Total number of purses is $na$ and the total number of coins is $mn$.
Now there are two specified coins. Let it be $A$ and $B$.
A purse contains $n$ coins. So if $A$ is placed into one purse, there remains $n - 1$ positions. At the same time, on the whole there remain $mn - 1$ coins.
So the chance of $B$with $A$ in the same purse is obtained by dividing the favourable by the total.
That is, the chance of $B$ with $A$ is equal to $\dfrac{{n - 1}}{{mn - 1}}$.
So $\dfrac{{n - 1}}{{mn - 1}}$ is the answer of (1).

We can see that the chance of $B$ not with A is equal to the difference of one and the chance of $B$ with $A$.
That is, the chance of $B$ not with A $ = 1 - (\dfrac{{n - 1}}{{mn - 1}}) = \dfrac{{mn - 1 - (n - 1)}}{{mn - 1}} = \dfrac{{mn - 1 - n + 1}}{{mn - 1}} = \dfrac{{n(m - 1)}}{{mn - 1}}$
Now we are asked to find the chance, when $r$ purses have been examined and found not to contain either of the specified coins.
For, consider the $m - r$ purses which have not been examined.
If $A$ and $B$ are together, the chance that they occur in these purses is obtained by dividing the number of purses not examined by $m$.
That is, the chance that they occur in these $m - r$ purses $ = \dfrac{{m - r}}{m}$.
 If $A$ and $B$ are apart, the chance that they occur in these purses is the number of ways in which they can occur separately in any two of the purses we consider divided by the total number of ways in which they can occur separately in any two purses.
Number of ways in which they can occur separately in any two of the $m - r$ purses we are considering is $(m - r)(m - r - 1)$.
Total number of ways in which they can occur separately in any two purses is \[m(m - 1)\].
Therefore, if $A$ and $B$ are apart, the chance that they occur in these purses $ = \dfrac{{(m - r)(m - r - 1)}}{{m(m - 1)}}$

So the chance when $r$ purses have been examined and found not to contain either of the specified coins is given by $\dfrac{{(\dfrac{{n - 1}}{{mn - 1}})(\dfrac{{m - r}}{m})}}{{(\dfrac{{n - 1}}{{mn - 1}})(\dfrac{{m - r}}{m}) + (\dfrac{{n(m - 1)}}{{mn - 1}})(\dfrac{{(m - r)(m - r - 1)}}{{m(m - 1)}})}}$
We can cancel $(\dfrac{{m - r}}{m})$ from each term we get $\dfrac{{(\dfrac{{n - 1}}{{mn - 1}})}}{{(\dfrac{{n - 1}}{{mn - 1}}) + (\dfrac{{n(m - 1)}}{{mn - 1}})(\dfrac{{m - r - 1}}{{m - 1}})}}$
Also we can cancel $\dfrac{1}{{mn - 1}}$ from each term gives $\dfrac{{n - 1}}{{n - 1 + n(m - 1)(\dfrac{{m - r - 1}}{{m - 1}})}}$
Simplifying we get $\dfrac{{n - 1}}{{n - 1 + n(m - r - 1)}} = \dfrac{{n - 1}}{{n - 1 + nm - nr - n}} = \dfrac{{n - 1}}{{nm - nr - 1}}$

$\therefore $ So the required chance in (2) is $\dfrac{{n - 1}}{{nm - nr - 1}}$.

Note:
Here in the question it is said about the total number of coins and coins in each bag. Using these values we had calculated the favourable number and total number in each case. If we take the given total number instead, it will lead to a wrong answer.