Question

The probability of a man hitting the target is $\frac{1}{4}$. The number of times atleast must he fire so that the probability of his hitting the target atleast once is greater than $\frac{2}{3}$ is

• A. 6
• B. 5
• C. 4
• D. 3

Answer 1

The correct option is C 4

Proboability of hitting target once $=$ $\frac{1}{4}$

$1-P\left(nothittingtarget\right)>\frac{2}{3}P\left(nothittingtarget\right)<\frac{1}{3}{\frac{3}{4}}^n<\frac{1}{3}$

Hence, $=4$