JAMB 1999 · UME · Q41

Probability

Three boys play a game of luck in which their respective chances of winning are $\frac12$ , $\frac13$ and $\frac14$ . What is the probability that one and only one of the boys wins the game?

Worked solution (try it first)

Each boy loses with probability $\frac12$ , $\frac23$ and $\frac34$ respectively.
"Only one wins" means one wins and the other two lose.
First only: $\frac12 \cdot \frac23 \cdot \frac34 = \frac{6}{24}$ .
Second only: $\frac12 \cdot \frac13 \cdot \frac34 = \frac{3}{24}$ .
Third only: $\frac12 \cdot \frac23 \cdot \frac14 = \frac{2}{24}$ .
These can't happen together, so add: $\frac{6 + 3 + 2}{24} = \frac{11}{24}$ , option C.

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