The probability of a man hitting a target is \[0.25\]. He shoots \[7\] times. What is the probability of his hitting at least twice?

Here, we have n = \[7\], \[p\text{ }=\text{ }0.25\text{ }=\text{ }25/100\text{ }=\text{ 1/4}\] and \[q=1-1/4=3/4\]

\[P\left( X\text{ }\ge \text{ }2 \right)\text{ }=\text{ }1\text{ }\text{ }\left[ P\left( X\text{ }=\text{ }0 \right)\text{ }+\text{ }P\left( X\text{ }=\text{ }1 \right) \right]\]

Therefore, the required probability is \[4547/8192\].