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%
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%