Minimum number of times a fair coin must be tossed, so that the probability of getting at least one head, is…
Minimum number of times a fair coin must be tossed, so that the probability of getting at least one head, is more than $99 \%$ is
5
6
7
8
Solution
Let the coin be tossed ' $n$ ' number of times.
Probability of getting head is $\mathrm{p}=\frac{1}{2}$
$\begin{array}{ll}
\therefore & \mathrm{q}=1-\frac{1}{2}=\frac{1}{2} \\
& \mathrm{P}(\mathrm{X} \geq 1)\gt\frac{99}{100} \\
\therefore & 1-\mathrm{P}(\mathrm{X}=0)\gt\frac{99}{100} \\
\therefore & 1-\left(\frac{1}{2}\right)^n\gt\frac{99}{100} \\
\therefore & \left(\frac{1}{2}\right)^n \lt \frac{1}{100} \\
\therefore & 100 \lt 2^n
\end{array}$
$\therefore \quad$ Minimum value of n is 7 .