Question

A bag contains four tickets numbered 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. $\frac{1}{64}$
• C. $\frac{5}{256}$
• D. $\frac{7}{256}$

Answer 1

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

The total number of ways of choosing the tickets is
$4\times 4\times 4\times 4=256$
The number of ways in which the sum can be 23.
= coefficients of $x^{23}$ in $(1+x+x^{10}+x^{11}{)}^4$
= coefficients of $x^{23}$ in $(1+x{)}^4(1+x^{10}{)}^4$
= coefficients of $x^{23}$ in $(1+4x+6x^2+4x^3+x^4)\times (1+4x^{10}+6x^{20})$
$=4\times 6=24$
The probability of required event =$\frac{24}{256}=\frac{3}{32}$