(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.