Question

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?

Answer 1

Let hitting the target be a success in a shoot.
We have,

\[p =\text{  probability of hitting the target } = 0 . 25 = \frac{1}{4}\]
\[\text{ Also,}  q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}\]
\[\text{ Let X denote the number of success in a sample of 7 trials . Then, } \]
\[\text{ X follows binomial distribution with parameters n } = 7 \text{ and } p = \frac{1}{4}\]
\[ \therefore P\left( X = r \right) = ^{7}{}{C}_r p^r q^\left( 7 - r \right) = ^{7}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^\left( 7 - r \right) = \frac{^{7}{}{C}_r 3^\left( 7 - r \right)}{4^7}, \text{ where r }  = 0, 1, 2, 3, 4, 5\]
\[\text{ Now,}  \]
\[\text{ Required probability } = P\left( X \geq 2 \right)\]
\[ = 1 - \left[ P\left( X = 0 \right) + P\left( X = 1 \right) \right]\]
\[ = 1 - \left[ \frac{^{7}{}{C}_0 3^7}{4^7} + \frac{^{7}{}{C}_1 3^6}{4^7} \right]\]
\[ = 1 - \left[ \frac{2187}{16384} + \frac{5103}{16384} \right]\]
\[ = 1 - \frac{7290}{16384}\]
\[ = \frac{9094}{16384}\]
\[ = \frac{4547}{8192}\]