Question

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

Answer 1

Suppose the man tosses a fair coin n times and X denotes the number of heads in n tosses.

\[\text{ As p } = \frac{1}{2} \text{ and q } = \frac{1}{2}, \]
\[ P(X = r) = ^{n}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{n - r} , r = 0, 1, 2, 3 . . . . n\]
\[\text{ It is given that } P\left( X \geq 1 \right) > 0 . 9\]
\[ \Rightarrow 1 - P\left( X = 0 \right) > 0 . 9\]
\[ \Rightarrow 1 - ^{n}{}{C}_0 \left( \frac{1}{2} \right)^n > 0 . 9\]
\[ \Rightarrow \left( \frac{1}{2} \right)^n < \frac{1}{10}\]
\[ \Rightarrow 2^n > 10\]
\[ \Rightarrow n = 4, 5, 6 . . . . \]
\[\text{ Hence, the man must toss the coin at least 4 times } . \]