The probability of a shooter of hitting the target is $\frac{1}{4}$. The minimum number of fire needed so that the probability of hitting the target atleast once is greater than $\frac{7}{16}$ is:

3

The correct answer is Option (3) → 3

Given:

Probability of hitting target in one shot: $p = \frac{1}{4}$

Probability of missing target in one shot: $q = 1 - p = \frac{3}{4}$

Let $n$ = number of shots. Probability of hitting target at least once:

$P(\text{at least one hit}) = 1 - q^n = 1 - \left(\frac{3}{4}\right)^n$

Set inequality:

$1 - \left(\frac{3}{4}\right)^n > \frac{7}{16}$

$\left(\frac{3}{4}\right)^n < 1 - \frac{7}{16} = \frac{9}{16}$

Check powers of $3/4$:

$\left(\frac{3}{4}\right)^1 = 3/4 = 12/16 > 9/16$ → not enough

$\left(\frac{3}{4}\right)^2 = 9/16 = 9/16$ → equal, need > → not enough

$\left(\frac{3}{4}\right)^3 = 27/64 \approx 0.4219 < 9/16 = 0.5625$ → satisfies

Hence, minimum $n = 3$