Four tickets are marked 00, 01, 10 and 11 respectively, are placed in a bag. A ticket is drawn at random five times, being replaced each time. The probability that the sum of the numbers on the five tickets drawn is 24, is

$\frac{25}{512}$

We have,

Number of ways of drawing five tickets = $4 ×4×4×4×4=4^5$

So, total number of elementary events = $4^5$

Now,

Favourable number of elementary events

= Number of ways of getting 24 as the sum of the numbers on the five tickets.

= Coefficient of $x^{24}$ in $(x^0+x^1 +x^{10}+x^{11})^5$

= Coefficient of $x^{24}$ in $(1 + x)^5 (1+x^{10})^5$

= Coefficient of $x^{24}$ in $(1 + x)^5 ({^5C}_0 + {^5C}_1 x^{10}+{^5C}_2 x^{20}+...)$

=$ {^5C}_2 $× Coefficient of $x^4$ in $(1 + x)^5={^5C}_2 × {^5C}_4=50$

Hence, required probability $=\frac{50}{4^5}=\frac{25}{512}$