Question

If $6n$ tickets numbered $0,1,2,....6n-1$ are placed in a bag, and three are drawn out, show that the chance that the sum of the numbers on them is equal to $6n$ is $\frac{3n}{(6n-1)(6n-2)}$.

Answer 1

The total no. of ways in which three tickets may be drawn in

$\frac{6n(6n-1)(6n-2)}{\mathrm{3.2.1}}=n(6n-1)(6n-2)$

To find the no. of ways in which the sum of the numbers drawn is $6n$ we may proceed as follows;

First suppose $0$ is drawn; then we have to make up $6n$ in all possible ways from two of tickets. $1,2,3,....6n-1$ this can be done in $3n-1$ ways

Then if $1$ is drawn, $6n-1$ from two of the tickets $2,3,...6n-1$, it can be done in $3n-2$ ways

If $2$ is drawn, we have to make up $6n-2$ from two of the numbers $3,4,,,...6n-1$; it can done in $3n-4$ ways. $3$ is drawn then there are $3n-5$ ways to make $6n-3$.

Hence the no. of ways of making up $6n$ in sum of $2n$ terms;-

$(3n-1)+(3n-2)+(3n-4)+(3n-5)+......+(5+4)+(2+1)$

$=6n-3+6n-9+6n-12+...$

$=3n^2$

$\therefore$ Required chance$=\frac{3n^2}{n(6n-1)(6n-2)}$

Or the Required chance$=\frac{3n}{(6n-1)(6n-2)}$