Question

If the probability distribution of a discrete random variable X is given by \( P(X=k) = \frac{2^{-k}(3k+1)}{2^c} \), k = 0, 1, 2, ..., \( \infty \) then P(X \( \le \) c) = (The expression seems to be \( \frac{2^{-k}(3k+1)}{K} \) where K is a constant, or \(2^c\) is part of the constant. Assuming \(2^c\) is the normalization constant \(K\).)

• A. \( \frac{c}{5} \)
• B. \( \frac{c}{4} \)
• C. \( \frac{c+2}{5} \)
• D. \( \frac{c-2}{7} \) Correct Answer

Hint:

For a discrete probability distribution \( P(X=k) \), the sum over all possible \( k \) must be 1: \[ \sum_k P(X=k) = 1 \] This helps find normalization constants.

Useful series sums for geometric and related series: \[ \sum_{k=0}^{\infty} x^k = \frac{1}{1-x} \quad \text{for} \quad |x| < 1 \] \[ \sum_{k=1}^{\infty} kx^{k-1} = \frac{1}{(1-x)^2} \quad \text{for} \quad |x| < 1 \] \[ \sum_{k=0}^{\infty} kx^k = x \sum_{k=1}^{\infty} kx^{k-1} = \frac{x}{(1-x)^2} \] \[ P(X \le c) = \sum_{k=0}^{c} P(X=k) \]

Answer 1

The Correct Option is B

The question seems to have a typo.
The sum of probabilities must be 1.
Let \( P(X=k) = C \cdot 2^{-k}(3k+1) \), where \( C = \frac{1}{2^c} \) is the normalization constant.
So, \( \sum_{k=0}^{\infty} C \cdot 2^{-k}(3k+1) = 1 \).
\( C \sum_{k=0}^{\infty} (3k+1) \left(\frac{1}{2}\right)^k = 1 \).
Let \( S = \sum_{k=0}^{\infty} (3k+1)x^k = 3x \sum_{k=1}^{\infty} k x^{k-1} + \sum_{k=0}^{\infty} x^k \), where \( x = \frac{1}{2} \).
We know \( \sum_{k=0}^{\infty} x^k = \frac{1}{1-x} \) for \( |x|<1 \).
And \( \sum_{k=1}^{\infty} kx^{k-1} = \frac{d}{dx} \left(\sum_{k=0}^{\infty} x^k\right) = \frac{d}{dx} \left(\frac{1}{1-x}\right) = \frac{1}{(1-x)^2} \).
So, \( S = 3x \cdot \frac{1}{(1-x)^2} + \frac{1}{1-x} \).
Substitute \( x=\frac{1}{2} \): \( 1-x = \frac{1}{2} \).
\( S = 3\left(\frac{1}{2}\right) \frac{1}{\left(\frac{1}{2}\right)^2} + \frac{1}{\frac{1}{2}} = \frac{3}{2} \cdot \frac{1}{1/4} + 2 = \frac{3}{2} \cdot 4 + 2 = 6+2 = 8 \).
So, \( C \cdot S = C \cdot 8 = 1 \implies C = \frac{1}{8} \).
Therefore, \( 2^c = 8 \implies c = 3 \).
(This assumes \(c\) is an integer related to the normalization).
Now we need to find \( P(X \le c) \).
If \(c=3\), we need \( P(X \le 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) \).
\( P(X=k) = \frac{1}{8} \left(\frac{1}{2}\right)^k (3k+1) \).
\( P(X=0) = \frac{1}{8} (1)(1) = \frac{1}{8} \).
\( P(X=1) = \frac{1}{8} \left(\frac{1}{2}\right)(3(1)+1) = \frac{1}{8} \cdot \frac{1}{2} \cdot 4 = \frac{4}{16} = \frac{1}{4} \).
\( P(X=2) = \frac{1}{8} \left(\frac{1}{4}\right)(3(2)+1) = \frac{1}{8} \cdot \frac{1}{4} \cdot 7 = \frac{7}{32} \).
\( P(X=3) = \frac{1}{8} \left(\frac{1}{8}\right)(3(3)+1) = \frac{1}{8} \cdot \frac{1}{8} \cdot 10 = \frac{10}{64} = \frac{5}{32} \).
\( P(X \le 3) = \frac{1}{8} + \frac{1}{4} + \frac{7}{32} + \frac{5}{32} = \frac{4}{32} + \frac{8}{32} + \frac{7}{32} + \frac{5}{32} = \frac{4+8+7+5}{32} = \frac{24}{32} = \frac{3}{4} \).
If \(c=3\), then \( P(X \le c) = \frac{3}{4} \).
Let's check the options with \(c=3\):
 

• Option (1): \( \frac{c}{5} = \frac{3}{5} \)
• Option (2): \( \frac{c}{4} = \frac{3}{4} \)
• Option (3): \( \frac{c+2}{5} = \frac{3+2}{5} = \frac{5}{5} = 1 \)
• Option (4): \( \frac{c-2}{7} = \frac{3-2}{7} = \frac{1}{7} \)

Option (2) matches \( \frac{3}{4} \) when \( c = 3 \).
This interpretation seems consistent.
The parameter \(c\) in \(P(X \le c)\) is the same \(c\) as in \( 2^c \).