In a Binomial distribution, the probability of getting a success is $\frac{3}{4}$ and the variance is $\frac{3}{8}$ then the probability of no 8 success is:

$\frac{1}{16}$

The correct answer is Option (3) → $\frac{1}{16}$

Given: $p = \frac{3}{4}, \ q = 1-p = \frac{1}{4}$, Variance: $\sigma^2 = n p q = \frac{3}{8}$

$n \cdot \frac{3}{4} \cdot \frac{1}{4} = \frac{3}{8} \Rightarrow n \cdot \frac{3}{16} = \frac{3}{8} \Rightarrow n = 2$

Probability of no success (0 success) in $n=2$ trials:

$P(0 \ \text{success}) = q^n = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$

Answer: $\frac{1}{16}$