Discussion
If a random variable \(X\) assumes only positive integral values, with the probability
\[ P(X = x) = \frac{2}{3}\left(\frac{1}{3}\right)^{x-1}, \]
\(x = 1, 2, 3, \ldots,\)
then \(E(X)\) is
Answer: D
The expectation is 3/2.
\(P(X = x) = \frac{2}{3}\left(\frac{1}{3}\right)^{x-1}\) is the geometric distribution counting trials up to the first success, with success probability \(p = \frac{2}{3}\). Its mean is \(1/p\):
\[ \begin{aligned} E(X) &= \sum_{x \ge 1} x\cdot\tfrac{2}{3}\left(\tfrac{1}{3}\right)^{x-1} \\ &= \tfrac{2}{3}\cdot\frac{1}{(1 - 1/3)^{2}} = \tfrac{2}{3}\cdot\tfrac{9}{4} = \tfrac{3}{2} \end{aligned} \]
The option 2/3 is \(p\) itself, 2/9 is \(p(1-p)\), and 1 would mean the variable always equals 1. Since \(X \ge 1\) with positive probability of larger values, its mean must exceed 1, which already rules out three options.
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