A bag contains 4 white and 2 black balls. Two balls are drawn at random, one after the other, without replacement. If X denotes the number of black balls drawn, then \(E(X)\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{2}{3}\)
X can be 0, 1 or 2. Two balls out of 6 can be chosen in \({}^{6}C_{2} = 15\) ways:
\( \displaystyle P(X = 0) = \frac{{}^{4}C_{2}}{15} = \frac{6}{15} \)
\( \displaystyle P(X = 1) = \frac{{}^{4}C_{1} \times {}^{2}C_{1}}{15} = \frac{8}{15} \)
\( \displaystyle P(X = 2) = \frac{{}^{2}C_{2}}{15} = \frac{1}{15} \)
The expectation is \(\sum x\,P(X = x)\):
\( \displaystyle E(X) = 0 \times \frac{6}{15} + 1 \times \frac{8}{15} + 2 \times \frac{1}{15} \)
\( \displaystyle = \frac{10}{15} = \frac{2}{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{3}\) — Incorrect. \(\frac{1}{3}\) is the chance that a single drawn ball is black; two balls are drawn.
Option (2) \(\displaystyle\frac{8}{15}\) — Incorrect. This is \(P(X = 1)\) only; the term \(2 \times \frac{1}{15}\) for two black balls is missing.
Option (3) 1 — Incorrect. 1 is the middle value of X, not its mean; the values are not equally likely.