Question

This past summer the LifeSaving Society did a survey about who was at risk in the water, with the goal of reducing swimming accidents. (Globe and Mail July 15 2010)

Consider a similar study done that shows that 96 % of those born in Canada can swim but only 82 % of new Canadians can swim. In a random group of 9 new Canadians, find the probability that exactly 6 cannot swim.

Answer 1

prob that a new Canadian can swim = .82 let that be Y

prob that a new Canadian cannot swim = .18 let that be N

so we are looking at something like NNNNNNYYY
that particular prob would be (.18^6(.82)^3

but the NNNNNNYYY can be arranged in 9!/(6!3!) ways or C(9,6) or 84 ways

so prob of your event
= 84(.18^6)(.82^3) = appr. .00158

(notice that the fact that 96% of those born in Canada can swim is irrelevant to the question)