Question

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

• A. $\frac{3}{32}$
• B. $\frac{1}{64}$
• C. $\frac{2}{256}$
• D. $\frac{7}{256}$

Answer 1

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

$23$ is possible if the tickets are $11,11,01,00$ or $11,10,01,01$
Hence, probability $=(\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4})\times \frac{4!}{2!}+(\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4})\times \frac{4!}{2!}=\frac{3}{32}$
{The term $\frac{4!}{2!}$ is used to arrange the $4$ numbers in order where one number has repeated in either case}