Question

If mohan has $3$ tickets of a lottery containing 3 prizes and $9$ blanks, then his chance of winning prize are

• A. $34/55$
• B. $21/55$
• C. $17/55$
• D. None of these

Answer 1

The correct option is A $34/55$

$3$ prizes and $3$ blanks

His chances of winning will be

$\frac{3_{C_1}\times 9_{C_2}+3_{C_2}\times 9C_1+3_{C_3}\times 9_{C_0}}{{12}_{C_3}}$

$=\frac{3!}{2!1!}\times \frac{9!}{7!2!}$$+\frac{3!}{2!1!}\times \frac{9!}{8!1!}+1\times 1$

$=\frac{136}{220}=\frac{34}{55}$