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

4

The correct answer is Option (2) → 4

Let the number of tosses be $n$.

Probability of at least one head:

$1-(\frac{1}{2})^{n} > 0.9$

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

Take $\log$:

$n\log(\frac{1}{2}) < \log(0.1)$

$n > \frac{\log(0.1)}{\log(\frac{1}{2})}$

$n > \frac{-1}{-\log 2}$ (base 10 logs)

$n > \frac{1}{\log 2}$

$\log 2 \approx 0.3010$

$n > 3.32$

Minimum integer $n=4$.

Final answer: $4$ tosses