Question

Four tickets marked $00,01,10,11$ 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 number on tickets thus drawn is $23$, is

• A. $\frac{25}{256}$
• B. $\frac{100}{256}$
• C. $\frac{231}{256}$
• D. None of these

Answer 1

The correct option is A $\frac{25}{256}$

The total number of ways in which $4$ tickets can be drawn $5$ times $=4^5=1024$
Favorable number of ways of getting a sum of $23$
$=$ coefficients of $x^{23}$ in ${(x^{00}+x^{01}+x^{10}+x^{11})}^5$
$=$ coefficient of $x^{23}$ in ${\left((1+x)(1+x^{10})\right)}^5$
$=$ coefficients of $x^{23}$ in ${(1+x)}^5{(1+x^{10})}^5$
$=$ coefficients of $x^{23}$ in
$(1+5x+10x^2+10x^3+5x^4+x^5)(1+5x^{10}+10x^{20}+10x^{30}+...)=100$
$\therefore$ required probability $=\frac{100}{1024}=\frac{25}{256}$