Question

If the probability that random variable $X$ takes values $x$ is given by $P(X=x)=k(x+1)3^{-x}, x=0,1,2,3, \dots$, where $k$ is a constant, then $P(X \geq 2)$ is equal to

• A. $\frac{7}{27}$
• B. $\frac{11}{18}$
• C. $\frac{7}{18}$
• D. $\frac{20}{27}$

Answer 1

Answer: (A)

$\frac{7}{27}$

Answer 2

The sum of probabilities must be 1: $\sum_{x=0}^{\infty} P(X=x) = \sum_{x=0}^{\infty} k(x+1)3^{-x} = 1$. This sum evaluates to $k \cdot \frac{9}{4} = 1$, so $k = \frac{4}{9}$. The probability mass function is $P(X=x) = \frac{4}{9}(x+1)3^{-x}$. We need $P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X=0) + P(X=1))$. $P(X=0) = \frac{4}{9}$ and $P(X=1) = \frac{8}{27}$. Thus, $P(X < 2) = \frac{4}{9} + \frac{8}{27} = \frac{20}{27}$. Therefore, $P(X \geq 2) = 1 - \frac{20}{27} = \frac{7}{27}$.