If the mean and variance of binomial distribution are 2 and 1 respectively, then the probability of at least 1 success is:
Answer & explanation
Correct answer: option 4
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 $\frac{15}{16}$.