Question

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

• A. \( \frac{7}{27} \)
• B. \( \frac{4}{9} \)
• C. \( \frac{8}{27} \)
• D. \( \frac{1}{9} \)

Hint:

For a discrete probability distribution, the sum of probabilities over all possible values of the random variable must equal 1. Use this property to find the value of the constant \( k \). To calculate \( P(X \ge a) \), it is often easier to calculate \( 1 - P(X<a) = 1 - \sum_{x=0}^{a-1} P(X = x) \). Remember the formula for the sum of an infinite geometric series and its derivatives.

Answer 1

The Correct Option is D

Since \( P(X = x) \) defines a probability distribution, the sum of probabilities over all possible values of \( x \) must be equal to 1: \[ \sum_{x=0}^{\infty} P(X = x) = 1 \] \[ \sum_{x=0}^{\infty} k(x + 1)3^{-x} = 1 \] \[ k \sum_{x=0}^{\infty} (x + 1)\left(\frac{1}{3}\right)^x = 1 \] Let \( S = \sum_{x=0}^{\infty} (x + 1)\left(\frac{1}{3}\right)^x = 1 \cdot \left(\frac{1}{3}\right)^0 + 2 \cdot \left(\frac{1}{3}\right)^1 + 3 \cdot \left(\frac{1}{3}\right)^2 + 4 \cdot \left(\frac{1}{3}\right)^3 + \dots \) \[ S = 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \dots \quad ...(i) \] Multiply by \( \frac{1}{3} \): \[ \frac{1}{3}S = \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \dots \quad ...(ii) \] Subtract (ii) from (i): \[ S - \frac{1}{3}S = \left(1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \dots\right) - \left(\frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \dots\right) \] \[ \frac{2}{3}S = 1 + \left(\frac{2}{3} - \frac{1}{3}\right) + \left(\frac{3}{9} - \frac{2}{9}\right) + \left(\frac{4}{27} - \frac{3}{27}\right) + \dots \] \[ \frac{2}{3}S = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots \] The right side is a geometric series with first term \( a = 1 \) and common ratio \( r = \frac{1}{3} \). The sum is \( \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \). \[ \frac{2}{3}S = \frac{3}{2} \] \[ S = \frac{3}{2} \times \frac{3}{2} = \frac{9}{4} \] So, \( kS = 1 \implies k \cdot \frac{9}{4} = 1 \implies k = \frac{4}{9} \). Now we need to find \( P(X \ge 3) = 1 - P(X<3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] \). \[ P(X = 0) = k(0 + 1)3^{-0} = \frac{4}{9} \cdot 1 \cdot 1 = \frac{4}{9} \] \[ P(X = 1) = k(1 + 1)3^{-1} = \frac{4}{9} \cdot 2 \cdot \frac{1}{3} = \frac{8}{27} \] \[ P(X = 2) = k(2 + 1)3^{-2} = \frac{4}{9} \cdot 3 \cdot \frac{1}{9} = \frac{12}{81} = \frac{4}{27} \] \[ P(X \ge 3) = 1 - \left( \frac{4}{9} + \frac{8}{27} + \frac{4}{27} \right) = 1 - \left( \frac{12}{27} + \frac{8}{27} + \frac{4}{27} \right) = 1 - \frac{24}{27} = 1 - \frac{8}{9} = \frac{1}{9} \]

Answer 2

To solve the problem, we need to find the probability \( P(X \ge 3) \) where the probability function is given by \(P(X = x) = k(x + 1)3^{-x}\) for \( x = 0, 1, 2, 3, \ldots \), and \( k \) is a constant.

Step 1: Determine the value of the constant \( k \)

Since \( P(X = x) \) is a probability function, the sum of all probabilities must be equal to 1:

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

This means:

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

First, let's calculate \(\sum_{x=0}^{\infty} (x + 1)3^{-x}\) using the properties of geometric series and their derivatives:

Consider the series \(\sum_{x=0}^{\infty} 3^{-x} = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}\).

Now, consider \(f(t) = \sum_{x=0}^{\infty} t^x = \frac{1}{1 - t}\).

Differentiate with respect to \(t\):

\(\sum_{x=0}^{\infty} xt^{x-1} = \frac{1}{(1 - t)^2}\).

By multiplying through by \(t\), we have:

\(\sum_{x=0}^{\infty} xt^x = \frac{t}{(1 - t)^2}\).

For our series, substitute \( t = \frac{1}{3} \):

\(\sum_{x=0}^{\infty} x \left( \frac{1}{3} \right)^x = \frac{\frac{1}{3}}{\left(1 - \frac{1}{3}\right)^2} = \frac{\frac{1}{3}}{\left(\frac{2}{3}\right)^2} = \frac{3}{4}\).

Thus, the series becomes:

\(\sum_{x=0}^{\infty} (x+1)\left(\frac{1}{3}\right)^x = \left(\frac{3}{2}\right) + \left(\frac{3}{4}\right) = \frac{9}{4}\).

Therefore, we have:

\(k \cdot \frac{9}{4} = 1\)

This implies that:

\(k = \frac{4}{9}\)

Step 2: Calculate \( P(X \ge 3) \)

We now need to calculate \( P(X \ge 3) = 1 - P(X < 3) \).

Calculate \( P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \).

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

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

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

Thus, \(P(X < 3) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27} = \frac{12}{27} + \frac{8}{27} + \frac{4}{27} = \frac{24}{27}\)

Therefore, \(P(X \ge 3) = 1 - \frac{24}{27} = \frac{3}{27} = \frac{1}{9}\).

Correct Answer: \(\frac{1}{9}\)