Assume that 10% of the population are le ft-handers. Wha t is the probability that in a class of 40 there are at least three left-handers?

This is equal to... 1 - P(less than 3 left-handed people)  ....so....

P(no left-handed people]  = C(40, 0)(.90)^40 = .0148

P(1 left-handed person) = C(40, 1) (.90)^39 (.10)^1 = .066

P(2 left-handed people)  = C(40, 2) (.90)^38 (.10)^2 = .142

So

P(at least 3 left-handed people)  = 1 - P(less than 3 left-handed people) =

1 - .0148 - .066 - .142  =  about 77.7%

Yes that is how I would have done it too. :)

Thanks, Melody.....!!!!