23. Tom and Geri have a competition. Initially, each player has one attempt at hitting a target. If one player hits the target and the other does not then the successful player wins. If both players hit the target, or if both players miss the target, then each has another attempt, with the same rules applying. If the probability of Tom hitting the target is always $\frac{4}{5}$ and the probability of Geri hitting the target is always $\frac{2}{3}$, what is the probability that Tom wins the competition?
A $\frac{4}{15}$
B $\frac{8}{15}$
C $\frac{2}{3}$
D $\frac{4}{5}$
E $\frac{13}{15}$
Solution: $\mathbf{C}$
Let the probability that Tom wins the competition be $p$. The probability that initially Tom hits and Geri misses is $\frac{4}{5} \times \frac{1}{3}=\frac{4}{15}$. The probability that initially they both hit is $\frac{4}{5} \times \frac{2}{3}=\frac{8}{15}$ and that they both miss is $\frac{1}{5} \times \frac{1}{3}=\frac{1}{15}$. So the probability that either they both hit or both miss is $\frac{8}{15}+\frac{1}{15}=\frac{9}{15}=\frac{3}{5}$. If they both hit or both miss the competition is in the same position as it was initially. So Tom's probability of winning is then $p$. Therefore, $p=\frac{4}{15}+\frac{3}{5} p$. So $\frac{2}{5} p=\frac{4}{15}$ and hence $p=\frac{2}{3}$.