Question

The probability of a man hitting a target is 1/4. If he fires 7 times, what is the probability of his hitting the target at least twice?

Answer 1

Let X be number of times the target is hit. Then, X follows a binomial distribution with n =7, \[p = \frac{1}{4}\text{ and }  q = \frac{3}{4}\]

\[P(X = r) = ^{7}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{7 - r} \]
\[P( \text{ hitting the target at least twice} )\]
\[ = P(X \geq 2) \]
\[ = 1 - \left\{ P(X = 0) + P(X = 1) \right\}\]
\[ = 1 -^{7}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{7 - 0} - ^{7}{}{C}_1 \left( \frac{1}{4} \right)^1 \left( \frac{3}{4} \right)^{7 - 1} \]
\[ = 1 - \left( \frac{3}{4} \right)^7 - 7\left( \frac{1}{4} \right) \left( \frac{3}{4} \right)^6 \]
\[ = 1 - \frac{1}{16384}(2187 + 5103) \]
\[ = 1 - \frac{3645}{8192}\]
\[ = \frac{4547}{8192}\]