(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?
Explanation
(a) \(p(Kunle) = \frac{1}{3} ; p(\text{not Kunle}) = \frac{2}{3}\)
\(p(Tayo) = \frac{1}{5} ; p(\text{not Tayo}) = \frac{4}{5}\)
\(p(\text{only one solve the question}) = p(\text{Kunle and not Tayo}) + p(\text{Tayo and not Kunle})\)
= \((\frac{1}{3} \times \frac{4}{5}) + (\frac{1}{5} \times \frac{2}{3})\)
= \(\frac{4}{15} + \frac{2}{15}\)
= \(\frac{6}{15} = \frac{2}{5}\)
(b) 10 persons to choose 8
Number of ways = \(^{10}C_{8}\)
= \(\frac{10!}{(10 - 8)! 8!}\)
= \(\frac{10 \times 9}{2}\)
= 45 ways.