Approximately 8.33% of men are colorblind. You survey 8 men from the population of a large city and count the number who are colorblind.

Question

This question shows up in the binomial distribution chapter in my book.

g. How many men would you have to survey in order to be at least 95% sure you would find at least $1$ who is colorblind?

Now I know that $P(X \geq 1) = 1 - P(X=0) = 1 - {8\choose0}.0833^0.9167^8 = 1 - .4986752922 = .5013247078$

But this is for a sample of 8 men. This question is asking me to find $n$ with $95$% confidence,

But this seems like a confidence interval question which I'm a little confused about since my book is about probability and not about statistics. Nothing like this shows up in my book, so I'm stuck. I feel like this is a "challenge" question.

Can someone give me a clue on what formula to use to deal with a question like this? Is this even a confidence interval question?

Thank You

Answer 1

You would have to solve $1 - {n\choose0}(0.0833^0)(0.9167^n)\ge0.95$.