Question

A bag contains four tickets $00,01,10,11$. Four tickets are chosen at random with replacement, the probability that sum of the numbers on the tickets is $23$ is

• A. $\frac{3}{32}$
• B. $\frac{1}{32}$
• C. $\frac{4}{32}$
• D. $\frac{5}{32}$

Answer 1

The correct option is A $\frac{3}{32}$

Total number of ways of choosing ticket $=4^4=256$

Number of ways in which sum can be $23$ $=$ coefficient of $x^{23}$ in $(x^0+x^1+x^{10}+x^{11}{)}^4$

$=$coefficient of $x^{23}$ in ${(1+x)}^4{(1+x^{10})}^4$

$=$coefficient of $x^{23}$ in $(1+4x+6x^2+4x^3+x^4)(1+4x^{10}+6x^{20}.....)$

$=4\times 6=24$

So desired probability $=\frac{24}{256}=\frac{3}{32}$

Hence, option A is correct.