JAMB Mathematics 2012 Paper

Concept:This is a binomial probability problem involving three independent tests, where we need exactly two successes (passes).Explanation:Let passing a test be $P$, with probability $P(P)=\frac{2}{3}$.Let failing a test be $F$, with probability $P(F)=1-\frac{2}{3}=\frac{1}{3}$.For exactly two passes and one failure, one possible outcome is $P, P, F$.Its probability is $\frac{2}{3}\times\frac{2}{3}\times\frac{1}{3}=\frac{4}{27}$.There are $3$ possible arrangements for two passes and one failure: $PPF$, $PFP$, and $FPP$.Thus, multiply by $3$:$$3 \times \frac{4}{27}=\frac{12}{27}=\frac{4}{9}.$$So the probability that he passes exactly two of the three tests is $\frac{4}{9}$.Answer:$\frac{4}{9}$, which is not correctly listed among the given options.