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

3

The correct answer is Option (2) → 3

When a fair coin is tossed, probability of getting a head = $p =\frac{1}{2}$,

So, $q = 1-\frac{1}{2}=\frac{1}{2}$

Let the number of trials be n.

Thus, we have a binomial distribution with $p =\frac{1}{2},q =\frac{1}{2}$ and number of trials = $n$.

∴ Probability of getting atleast one head = $P(X ≥ 1)$

$= 1 – P(0) = 1 – {^nC}_0q^n$

$= 1-1. (\frac{1}{2})^n = 1 -\frac{1}{2^n}$

According to given,

$1-\frac{1}{2^n}> 80\%⇒ 1 -\frac{1}{2^n}>\frac{80}{100}$

$⇒1-\frac{1}{2^n}>\frac{4}{5}⇒1-\frac{4}{5}>\frac{1}{2^n}$

$⇒\frac{1}{5}>\frac{1}{2^n}⇒5 < 2^n$

$⇒2^n>5$, which is satisfied if n is atleast 3.

Hence, the man must toss the coin atleast 3 times.