HMMT-Feb 2014 comb, Problem 3

Bob writes a random string of 5 letters, where each letter is either $A, B, C$, or $D$. The letter in each position is independently chosen, and each of the letters $A, B, C, D$ is chosen with equal probability. Given that there are at least two $A$ 's in the string, find the probability that there are at least three $A$ 's in the string.

Solution: There are $\left(\begin{array}{l}5 \\ 2\end{array}\right) 3^{3}=270$ strings with $2 A$ 's. There are $\left(\begin{array}{l}5 \\ 3\end{array}\right) 3^{2}=90$ strings with $3 A$ 's. There are $\left(\begin{array}{l}5 \\ 4\end{array}\right) 3^{1}=15$ strings with $4 A$ 's. There is $\left(\begin{array}{l}5 \\ 5\end{array}\right) 3^{0}=1$ string with $5 A$ 's. The desired probability is $\frac{90+15+1}{270+90+15+1}=\frac{53}{188}$. $\fbox{\overline{\frac{53}{188}}}$.

Answer: \overline{\frac{53}{188}}