If the mean and variance of binomial distribution are 2 and 1 respectively, then the probability of at least 1 success is:

$\frac{15}{16}$

The correct answer is Option (4) → $\frac{15}{16}$

$\text{Mean} = np = 2,\;\; \text{Variance} = npq = 1$

$npq = 1 \Rightarrow 2q = 1 \Rightarrow q=\frac{1}{2},\;\; p=\frac{1}{2}$

$np=2 \Rightarrow n\cdot\frac{1}{2}=2 \Rightarrow n=4$

$P(\text{at least 1 success}) = 1 - P(0)$

$= 1 - (1-p)^n = 1 - \left(\frac{1}{2}\right)^4$

$= 1 - \frac{1}{16} = \frac{15}{16}$

The probability is $\frac{15}{16}$.