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

3

The correct answer is Option (3) → 3

$P(\text{at least one head}) = 1 - P(\text{no head})$

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

$1 - \left(\frac{1}{2}\right)^n > 0.8$

$\left(\frac{1}{2}\right)^n < 0.2$

$n \log\left(\frac{1}{2}\right) < \log(0.2)$

$n > \frac{\log(0.2)}{\log(1/2)} \approx 2.32$

$\text{Smallest integer } n = 3$

$\text{Required number of tosses} = 3$