Question

What is the minimum number of times must a man toss a fair coin so that the probability of having at least one head is more than $90\%$ ?

Answer 1

Let the man toss the coin $n$ times. The $n$ tosses are $n$ Bernoulli trials.
Probability $\left(p\right)$ of getting a head at the toss of a coin is $\frac{1}{2}$
$p=\frac{1}{2},q=\frac{1}{2}$
$\therefore P(X=x){=}^n{C}_x{p}^{n-x}{q}^x{=}^n{C}_x{\left(\frac{1}{2}\right)}^{n-x}{\left(\frac{1}{2}\right)}^x{=}^n{C}_x{\left(\frac{1}{2}\right)}^n$
It is given that,
$P\left(\text{getting at least one head}\right)>\frac{90}{100}$
$P(X\ge 1)>0.9$
$\Rightarrow 1-P(X=0)>0.9$
$1{-}^n{C}_0\cdot \frac{1}{2^n}>0.9$
${}^n{C}_0\cdot \frac{1}{2^n}<0.1$
$\frac{1}{2^n}<0.1$
$2^n>\frac{1}{0.1}$
$2^n>10$ ....(1)
The minimum value of $n$ that satisfies the given inequality is $4$.
Thus, the man should toss the coin $4$ or more than $4$ times.