How to find a method of finding smallest sample size when given probabilities?

Question

Given this problem:

A random sample of n people is taken. Find the smallest possible value of n if the probability that none of these people takes more than t minutes to cook an egg is less than 0.003.

I also know that:

12% of people take more than t minutes to cook their egg.

So logically $P_{sample-100-people} = 0.88$

(where P is probability)

and I have to find $P_{sample-n-people} > 0.003$ .

The methord I am meant to use finds this by

$ 0.88^n < 0.003 $

From this point on the calculation is easy to do but I don't understand why one puts 0.88 to the nth power.

What is the logic, or reason behind this methord?

Answer 1

The probability that $k$ out of $n$ people take more than t minutes to cook their eggs is

$P(X=k)=\binom{n}{k} \cdot 0.12^k\cdot (1-0.12)^{n-k}$.

So $X$ is binomial distributed.

For $k=0$ we have

$P(X=0)=\binom{n}{0} \cdot 0.12^0\cdot (1-0.12)^{n-0}$

$P(X=0)=1 \cdot 1\cdot (1-0.12)^{n}$

$P(X=0)= 0.88^{n}$

The probability has to be less than $0.003=0.3\%$. I assume you mean that and not $0.003\%$. Thus the inequality is

$P(X=0)= 0.88^{n}<0.003$