Question

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

Answer 1

Let X be the number of heads and n be the minimum number of times that a man must toss a fair coin so that probability of X ≥ 1 is more than 80 % and X follows a binomial distribution with \[p = \frac{1}{2}, q = \frac{1}{2}\]
\[P(X = r) =^{n}{}{C}_r \left( \frac{1}{2} \right)^n \]
\[\text{ We have } P(X \geq 1) = 1 - P(X = 0) = 1 - ^{n}{}{C}_0 \left( \frac{1}{2} \right)^n = 1 - \frac{1}{2^n}\]
\[\text{ and } P(X \geq 1) > 80 \] %
\[1 - \frac{1}{2^n} > 80 \] % = 0 . 80 
\[\frac{1}{2^n} < 1 - 0 . 80 = 0 . 20\]
\[ 2^n > \frac{1}{0 . 2} = 5; \]
\[\text{ We know,}  2^2 < 5 \text{ while } 2^3 > 5\]
\[\text{ So, n } = 3 \]
\[\text{ So, n should be atleast } 3 .\]