If the mean and variance of a 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) → \(\displaystyle\frac{15}{16}\)
For a binomial distribution, mean = np and variance = npq:
\(np = 2\)
\(npq = 1\)
Dividing, \(q = \frac{1}{2}\), so \(p = 1 - q = \frac{1}{2}\) and \(n = \frac{2}{p} = 4\).
“At least 1 success” is the complement of “no success”:
\( \displaystyle P(X \ge 1) = 1 - P(X = 0) = 1 - q^4 = 1 - \left(\frac{1}{2}\right)^4 = \frac{15}{16} \)