Question

How many times must a man toss a fair coin so that the probability of having atleast one head is more than $90\%$?

Answer 1

Let the man tosses the coin n times.

Probability (p) of getting a head at the toss of a coin $=\frac{1}{2}$

$\frac{1}{2}$

and $q=1-\frac{1}{2}=\frac{1}{2}$

$\therefore P(X=r){=}^n{C}_r{p}^r{q}^{n-r}$

${=}^n{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{n-r}{=}^n{C}_r{\left(\frac{1}{2}\right)}^n$

It is given that

P(atleast one head) >$90\%$

$\Rightarrow 1-P\left(0\right)>\frac{90}{100}\Rightarrow 1{-}^n{C}_0{p}^0{q}^n>\frac{90}{100}\Rightarrow 1{-}^n{C}_0{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^n>\frac{9}{10}\Rightarrow 1-\frac{9}{10}>\frac{1}{2^n}\Rightarrow 2^n>10$

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.