Question

If X is a random variable with probability distribution $P(X=k) = \frac{(2k+3)c}{3^k}$, $k=0,1,2,\dots,\infty$, then $P(X=3) =$

• A. $\frac{1}{24}$
• B. $\frac{1}{18}$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Hint:

Recognizing arithmetico-geometric series is key for problems involving probability distributions of the form $(ak+b)r^k$. Remember the standard formulas for the sum of a geometric series $\sum x^k = \frac{1}{1-x}$ and the related series $\sum kx^k = \frac{x}{(1-x)^2}$ for $|x|<1$.

Answer 1

The Correct Option is B

Step 1: Determine constant c: The sum of probabilities must be 1. \[ \sum_{k=0}^{\infty} P(X=k) = 1 \implies c \sum_{k=0}^{\infty} \frac{2k+3}{3^k} = 1 \] Let \( S = \sum_{k=0}^{\infty} \frac{2k+3}{3^k} = 3 + \frac{5}{3} + \frac{7}{9} + \frac{9}{27} + \dots \) This is an Arithmetico-Geometric Series (AGP) with \( a=3, d=2, r=1/3 \). Multiply \( S \) by \( 1/3 \): \[ \frac{1}{3}S = 1 + \frac{5}{9} + \frac{7}{27} + \dots \] Subtract: \[ S - \frac{1}{3}S = 3 + \left(\frac{5}{3}-1\right) + \left(\frac{7}{9}-\frac{5}{9}\right) + \dots \] \[ \frac{2}{3}S = 3 + \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \dots \] \[ \frac{2}{3}S = 3 + 2\left(\frac{1}{3} + \frac{1}{9} + \dots\right) \] The term inside brackets is a GP with sum \( \frac{1/3}{1-1/3} = \frac{1/3}{2/3} = \frac{1}{2} \). \[ \frac{2}{3}S = 3 + 2\left(\frac{1}{2}\right) = 3 + 1 = 4 \] \[ S = 4 \times \frac{3}{2} = 6 \] Thus, \( 6c = 1 \implies c = \frac{1}{6} \).
Step 2: Calculate \( P(X=3) \): \[ P(X=3) = \frac{(2(3)+3)c}{3^3} = \frac{9c}{27} = \frac{c}{3} \] Substitute \( c = 1/6 \): \[ P(X=3) = \frac{1/6}{3} = \frac{1}{18} \]