Question

A has $3$ shares in a lottery in which there are $3$ prizes and $6$ blanks$;$ B has $1$ share in a lottery in which there is $1$ prize and $2$ blanks$:$ show that A's chance of success is to B's as $16$ to $7$.

Answer 1

A may draw $3$ prizes in $1$ way$;$
He may draw $2$ prizes and $1$ blank in $\frac{3.2}{1.2}\times 6$ ways $;$
He may draw $1$ prize and $2$ blanks in $3\times \frac{6.5}{1.2}$ ways$;$
The sum of these numbers is $64,$ which is the number of ways in which A can win a prize. Also he can draw $3$ tickets in $\frac{\mathrm{9.8.7}}{\mathrm{1.2.3}},$ or $84$ ways$;$
Therefore A's chance of success $=\frac{64}{84}=\frac{16}{21}$.
B's chance of success is clearly $=\frac{1}{3};$
Therefore A's chance $:$ B's chance $=\frac{16}{21}:\frac{1}{3}$
$=16:7$.