A variate X follows a Binomial distribution with probability of success 1/3 such that the probability of no success is \( \left(\displaystyle\frac{2}{3}\right)^{12} \), then the variance of this distribution is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{8}{3} \)
For a binomial distribution B(n, p), \( P(X=0)=q^n \).
\( \displaystyle p=\frac{1}{3},\quad q=1-p=\frac{2}{3} \)
\( \displaystyle q^n=\left(\frac{2}{3}\right)^{12}\ \Rightarrow\ n=12 \)
\( \displaystyle \text{Variance}=npq=12\times\frac{1}{3}\times\frac{2}{3}=\frac{8}{3} \)
Why the other options are incorrect
Option (1) 9 — Incorrect. The variance is \( \frac{8}{3} \), not 9; the variance npq is always less than the mean np = 4.
Option (2) \( \displaystyle\frac{2}{3} \) — Incorrect. \( \frac{2}{3} \) is q, the probability of failure, not the variance.
Option (4) 4 — Incorrect. 4 is the mean, np, not the variance npq.