The Probability mass functions of Random variable X is : \(P(X = x) = (0.6)^x(0.4)^{1-x}\); x = 0, 1 The variance of X is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.240
Putting x = 0 and x = 1:
\(P(X = 0) = 0.4,\quad P(X = 1) = 0.6\)
This is a Bernoulli distribution with p = 0.6. Using \(\mathrm{Var}(X) = E(X^2) - [E(X)]^2\):
\(E(X) = 0 \times 0.4 + 1 \times 0.6 = 0.6\)
\(E(X^2) = 0^2 \times 0.4 + 1^2 \times 0.6 = 0.6\)
\(\mathrm{Var}(X) = 0.6 - 0.36 = 0.24\)
This agrees with the Bernoulli result Var(X) = pq = 0.6 × 0.4 = 0.24.
Why the other options are incorrect
Option (1) 0.60 — Incorrect. 0.6 is the mean E(X), not the variance.
Option (2) 0.124 — Incorrect. The variance is 0.24.
Option (3) 0.244 — Incorrect. The variance is 0.24.