(a) The probability that Kunle solves a particular question is \(\frac{1}{3}\) while that of Tayo is \(\frac{1}{5}\). If both of them attempt the question, ...

Assessment: WAEC SSCE - Further Mathematics - 2018 (Essay) Subject: Further Mathematics

Question 1 Report

(a) The probability that Kunle solves a particular question is \(\frac{1}{3}\) while that of Tayo is \(\frac{1}{5}\). If both of them attempt the question, find the probability that only one of them will solve the question.

(b) A committee of 8 is to be chosen from 10 persons. In how many ways can this be done if there is no restriction?

Answer Details

(a) \(P(K)=\tfrac{1}{3}\Rightarrow P(K')=\tfrac{2}{3}\); \(P(T)=\tfrac{1}{5}\Rightarrow P(T')=\tfrac{4}{5}\). The events are independent.

Only one solves = (Kunle solves, Tayo fails) OR (Kunle fails, Tayo solves):

\[P=\left(\tfrac{1}{3}\times\tfrac{4}{5}\right)+\left(\tfrac{2}{3}\times\tfrac{1}{5}\right)=\frac{4}{15}+\frac{2}{15}=\frac{6}{15}=\frac{2}{5}\]

(b) Choosing 8 from 10 with no restriction:

\[\binom{10}{8}=\binom{10}{2}=\frac{10\times9}{2}=45\ \text{ways}\]

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