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%?
Answer & explanation
Correct answer: option 2
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