Question

A can hit a target 3 times in 6 shots, B : 2 times in 6 shots and C : 4 times in 4 shots. They fix a volley. What is the probability that at least 2 shots hit?

 

Answer 1

\[P\left( \text{ A hits the target }  \right) = \frac{3}{6}\]

\[P\left( \text{ B hits the target }  \right) = \frac{2}{6}\]

\[P\left( \text{ C hits the target }  \right) = \frac{4}{4} = 1\]

\[P\left( \text{ atleast 2 shots hit } \ \right) = P\left( \text{ exactly 2 shots hit }  \right) + P\left( \text{ all 3 shots hit}  \right)\]

\[ = \frac{3}{6}\left( 1 - \frac{2}{6} \right) + \frac{2}{6}\left( 1 - \frac{3}{6} \right) + \frac{3}{6} \times \frac{2}{6} \times 1 \text{( Here, the probability of C hitting the target is 1 . So, it will always hit ).} \]

\[\text{ When exactly 2 shots are hit, then either A hits or B hits } . \]

\[ = \frac{3}{6} \times \frac{4}{6} + \frac{2}{6} \times \frac{3}{6} + \frac{6}{36}\]

\[ = \frac{12 + 6 + 6}{36}\]

\[ = \frac{24}{36}\]

\[ = \frac{2}{3}\]