Four tickets 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 numbers on the ticket is ’15’ is

Let ‘$S$’ be the sample space and $E$ be the required event.

Now $n\left(s\right)=$ Total number of cases $=4^5=1024$ and

$n\left(E\right)=$ coefficient of  $x^{15}$ in  ${(x^0+x+x^{10}+x^{11})}^5$

$=$ coefficient of $x^{15}$ in $\left[{(1+x)}^5{(1+x^{10})}^5\right]$

$=$ coefficient of  $x^{15}$ in $(1+5x+10x^2+10x^3+5x^4+x^5)\times (1+5x^{10}+10x^{20}+\cdots )=5$

$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{5}{1024}$