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: Understanding the Concept
For any probability distribution, the sum of all probabilities must be equal to 1. We can use this property to find the value of the constant $c$. Once $c$ is known, we can calculate the probability for any specific value of the random variable, in this case, $P(X=3)$. 
Step 2: Key Formula or Approach 
1. Set up the summation: $\sum_{k=0}^{\infty} P(X=k) = 1$. 2. The sum is $\sum_{k=0}^{\infty} \frac{(2k+3)c}{3^k} = c \sum_{k=0}^{\infty} (2k+3) \left(\frac{1}{3}\right)^k = 1$. 3. We need to evaluate the infinite series. This is an arithmetico-geometric series. 4. Solve for $c$. 5. Calculate $P(X=3) = \frac{(2(3)+3)c}{3^3}$. 
Step 3: Detailed Explanation  
1. Evaluate the infinite series: Let $S = \sum_{k=0}^{\infty} (2k+3)x^k$ where $x = 1/3$. We can split the sum: \[ S = 2 \sum_{k=0}^{\infty} kx^k + 3 \sum_{k=0}^{\infty} x^k \] The second part is a standard geometric series: $\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$ for $|x|<1$. For $x=1/3$, this sum is $\frac{1}{1-1/3} = \frac{1}{2/3} = \frac{3}{2}$. The first part involves the sum $\sum_{k=0}^{\infty} kx^k = 0 \cdot x^0 + 1 \cdot x^1 + 2 \cdot x^2 + \dots = x(1+2x+3x^2+\dots)$. The series $1+2x+3x^2+\dots$ is the derivative of the geometric series $\sum x^k$, which is $(1-x)^{-2}$. So, $\sum_{k=1}^{\infty} kx^{k-1} = \frac{1}{(1-x)^2}$. Multiplying by $x$ gives $\sum_{k=1}^{\infty} kx^k = \frac{x}{(1-x)^2}$. For $x=1/3$, this sum is $\frac{1/3}{(1-1/3)^2} = \frac{1/3}{(2/3)^2} = \frac{1/3}{4/9} = \frac{1}{3} \times \frac{9}{4} = \frac{3}{4}$. Now, substitute back into the expression for $S$: \[ S = 2 \left(\frac{3}{4}\right) + 3 \left(\frac{3}{2}\right) = \frac{3}{2} + \frac{9}{2} = \frac{12}{2} = 6 \] 2. Solve for c: The total probability is 1: \[ \sum_{k=0}^{\infty} P(X=k) = c \cdot S = 1 \] \[ c \cdot 6 = 1 \implies c = \frac{1}{6} \] 3. Calculate P(X=3): Now we use the value of $c$ to find the required probability. \[ P(X=3) = \frac{(2(3)+3)c}{3^3} = \frac{(6+3) \cdot (1/6)}{27} \] \[ P(X=3) = \frac{9 \cdot (1/6)}{27} = \frac{9/6}{27} = \frac{3/2}{27} = \frac{3}{2 \times 27} = \frac{1}{2 \times 9} = \frac{1}{18} \] Step 4: Final Answer 
The value of the constant $c$ is $1/6$. The probability $P(X=3)$ is $\frac{1}{18}$.