A random variable X takes the values −1, 0 and 1 only. If \(E(X) = 0.2\) and \(\mathrm{Var}(X) = 0.36\), then \(P(X = 0)\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.6
Let \(P(X = -1) = a\) and \(P(X = 1) = b\). Then
\(E(X) = -a + b = 0.2\)
\(E(X^{2}) = a + b\)
Using \(\mathrm{Var}(X) = E(X^{2}) - [E(X)]^{2}\):
\(0.36 = (a + b) - (0.2)^{2}\)
\(a + b = 0.36 + 0.04 = 0.4\)
The probabilities add up to 1:
\(P(X = 0) = 1 - (a + b)\)
\(= 1 - 0.4 = 0.6\)
(Solving also gives b = 0.3 and a = 0.1, both valid probabilities.)
Why the other options are incorrect
Option (1) 0.64 — Incorrect. This is 1 − 0.36, taking \(E(X^{2})\) equal to the variance; the term \([E(X)]^{2} = 0.04\) is left out.
Option (2) 0.8 — Incorrect. This is 1 − 0.2, which uses the mean in place of a + b.
Option (4) 0.44 — Incorrect. This is 1 − (0.36 + 0.2); the mean must be squared before it is added.