How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?
Question:

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

Solution:

Let the man toss the coin n times. The n tosses are n Bernoulli trials.

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

$\therefore p=\frac{1}{2} \Rightarrow q=\frac{1}{2}$

$\therefore \mathrm{P}(\mathrm{X}=x)={ }^{n} \mathrm{C}_{x} p^{n-x} q^{x}={ }^{n} \mathrm{C}_{x}\left(\frac{1}{2}\right)^{n-x}\left(\frac{1}{2}\right)^{x}={ }^{n} \mathrm{C}_{x}\left(\frac{1}{2}\right)^{n}$

It is given that,

$P$ (getting at least one head) $>\frac{90}{100}$

$P(x \geq 1)>0.9$

$\Rightarrow 1-P(x=0)>0.9$

$P(x \geq 1)>0.9$

$\Rightarrow 1-P(x=0)>0.9$

$1-{ }^{n} C_{0} \cdot \frac{1}{2^{n}}>0.9$

${ }^{n} C_{0} \cdot \frac{1}{2^{n}}<0.1$

$\frac{1}{2^{n}}<0.1$

$2^{n}>\frac{1}{0.1}$

$2^{n}>10$                   (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.