Question

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

• A. \(\frac{5}{18}\)
• B. \(\frac{10}{18}\)
• C. \(\frac{20}{27}\)
• D. \(\frac{7}{27}\)

Hint:

For probability distributions, ensure the sum of all probabilities equals 1. Use geometric series formulas to simplify summations efficiently.

Answer 1

The Correct Option is D

The total probability is:

\[ \sum_{x=0}^\infty P(X = x) = 1. \]

Substitute \( P(X = x) = k(x + 1) 3^{-x} \):

\[ k \sum_{x=0}^\infty (x + 1) 3^{-x} = 1. \]

Let:

\[ S = \sum_{x=0}^\infty (x + 1) 3^{-x}. \]

Split \( S \) into two components:

\[ S = \sum_{x=0}^\infty 3^{-x} + \sum_{x=1}^\infty x \cdot 3^{-x}. \]

1. For the first term:

The sum of a geometric series is:

\[ \sum_{x=0}^\infty 3^{-x} = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}. \]

2. For the second term:

Using the formula for a weighted geometric series:

\[ \sum_{x=1}^\infty x \cdot 3^{-x} = \frac{\frac{1}{3}}{\left(1 - \frac{1}{3}\right)^2} = \frac{\frac{1}{3}}{\left(\frac{2}{3}\right)^2} = \frac{3}{4}. \]

Thus:

\[ S = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}. \]

Equating to 1:

\[ k \cdot \frac{9}{4} = 1 \implies k = \frac{4}{9}. \]

Finding \( P(X \geq 2) \):

\[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1). \]

\[ P(X = 0) = \frac{4}{9} \cdot (0 + 1) \cdot 3^0 = \frac{4}{9}. \]

\[ P(X = 1) = \frac{4}{9} \cdot (1 + 1) \cdot 3^{-1} = \frac{4}{9} \cdot 2 \cdot \frac{1}{3} = \frac{8}{27}. \]

\[ P(X \geq 2) = 1 - \frac{4}{9} - \frac{8}{27} = \frac{27}{27} - \frac{12}{27} - \frac{8}{27} = \frac{7}{27}. \]