Question

The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least $0.96$, is

Answer 1

Let $n$ be the required number.
$1-P\left(\text{no head}\right)-P\left(\text{one head}\right)\ge 0.96\Rightarrow 1-{}^n{C}_0{\left(\frac{1}{2}\right)}^n-{}^n{C}_1{\left(\frac{1}{2}\right)}^n\ge 0.96\Rightarrow 1-{\left(\frac{1}{2}\right)}^n-n{\left(\frac{1}{2}\right)}^n\ge 0.96\Rightarrow {\left(\frac{1}{2}\right)}^n(1+n)\le 0.04\Rightarrow \frac{n+1}{2^n}\le \frac{1}{25}\Rightarrow \frac{2^n}{n+1}\ge 25\therefore \text{Minimum value of}n=8$