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

.: Let the coin be tossed $n$ times.

$p$ be probability of getting a head $= \cfrac{1}{2}$

$q$ be probability of getting no head $= 1 - \cfrac{1}{2} = \cfrac{1}{2}$

$P({\rm{ at least one head }}) = P(X \ge 1) = 1 - P(X = 0)$

$= 1{ - ^n}{C_0}{q^n}{p^0} = 1 - (1){\left( {\cfrac{1}{2}} \right)^n}(1) = 1 - {\left( {\cfrac{1}{2}} \right)^n}$

According to question, $1 - {\left( {\cfrac{1}{2}} \right)^n} > \cfrac{{90}}{{100}}$

$\Rightarrow$ $1 - {\left( {\cfrac{1}{2}} \right)^n} > 0.9 \Rightarrow {\left( {\cfrac{1}{2}} \right)^n} < 1 - 0.9$

$\Rightarrow$ ${\left( {\cfrac{1}{2}} \right)^n} < 0.1 \Rightarrow n \ge 4$