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 coin be tossed n times. The only way to get no head is a tail every time, so by the complement rule:
\( \displaystyle P(\text{at least one head}) = 1 - \left(\frac{1}{2}\right)^n \)
This must be more than 0.9:
\( \displaystyle 1 - \frac{1}{2^n} \gt 0.9 \)
\( \displaystyle \frac{1}{2^n} \lt \frac{1}{10} \)
\(2^n \gt 10\)
Since \(2^3 = 8\) and \(2^4 = 16\), the least n is 4. Check: with 3 tosses the probability is \(1 - \frac{1}{8} = 87.5\%\), not enough; with 4 tosses it is \(1 - \frac{1}{16} = 93.75\%\).
CBSE Applied Mathematics (Class XII):
Probability Distribution
