Question

In a shooting competition, the probability of a man hitting a target is $\frac{3}{5}$. If he fires $4$ times, what is the probability of hitting target.

(1) at least twice
(2) at most twice

Answer 1

Probability of hitting target = $\frac{3}{5}$

Probability of hitting target atleast twice = 1 - (Probability of hitting once + Probability of all misses) = $1-(4\left(\frac{3\left(2\right)\left(2\right)\left(2\right)}{625}\right)+\frac{16}{625})=1-\left(\frac{112}{625}\right)=\frac{533}{625}$

Probability of hitting target atmost twice = 1 - (Probability of hitting thrice + Probability of all successful hits) = $1-(4\left(\frac{2\left(3\right)\left(3\right)\left(3\right)}{625}\right)+\frac{81}{625})=1-\left(\frac{297}{625}\right)=\frac{322}{625}$