How many times must a man toss a fair coin so that the probability of having at least one head is more than 80%?

Explanation

Let `X` be the number of heads and `n` be the minimum number of times that a man must toss a fair coin so that probability of `X geq 1` is more than `80 %` and `X` follows a binomial distribution with `p=frac{1}{2}, q=frac{1}{2} P(X=r)={ }^{n} C_{r}(frac{1}{2})^{n}`

We have `P(X geq 1)=1-P(X=0)=1-^{n} C_{0}(frac{1}{2})^{n}=1-frac{1}{2^{n}} and P(X geq 1)>80 %`

`1-frac{1}{2^{n}}>80 %=0.80`

`frac{1}{2^{n}}<1-0.80 2^{n}>frac{1}{0.2}=5`

We know,` 2^{2}<5` while `2^{3}>5`

So, `{n}=3`

So, `n` should be atleast `3` .