Question

URGENT! In a school, 8% of students are left handed. If 50 classes of 20 students are randomly selected, what is the probability of 10 classes have no left handed students?

I know that this is a binomial distribution question, but I can't seem to do this question even though I can apply the equation to simpler questions. Help is very much appreciated. Thank you!

Answer 1

prob(left) = .08
prob(right) = .92

let's first do the prob that in a class of 20 nobody is left-handed
= C(20,0)(.08)^0 (.92)^10
= .434388.. (I stored it)
or prob(1 or more lefthanders) = 1-.4343...= .56561...


Now we want the prob that of 50 classes contain no left-handed students
= C(50,10) (.4343..)^10 (.56561..)^40
= .0003099

I assumed you wanted exactly 10 of the 50 classes to contain no left-handers

Answer 2

Yes, I want exactly 10 of the 50 classes to contain no left handers.

So, instead of using .08 as the p for the last part, I should use the probability found from no left handers in a class of 20 instead? The question did ask me to find this as well, although I didn't put it up here as I have no problem solving it.

Thank you for the time and effort used to answer this question. I really appreciate it.