Question

Let \( X_1, X_2, X_3 \) be i.i.d. discrete random variables with the probability mass function
\[ p(k) = \left( \frac{2}{3} \right)^{k-1} \left( \frac{1}{3} \right), \quad k = 1, 2, 3, \dots \] Let \( Y = X_1 + X_2 + X_3 \). Then \( P(Y \geq 5) \) equals

• A. \( \frac{1}{9} \)
• B. \( \frac{25}{27} \)
• C. \( \frac{2}{27} \)
• D. \( \frac{8}{9} \)

Hint:

For sums of i.i.d. random variables, use the CDF and properties of the distribution to calculate the desired probability.

Answer 1

The Correct Option is D

Step 1: Understand the probability mass function (pmf).
The given pmf is: \[ p(k) = \left( \frac{2}{3} \right)^{k-1} \left( \frac{1}{3} \right), \quad k = 1, 2, 3, \dots \] This is the pmf of a geometric distribution with parameter \( p = \frac{1}{3} \), where \( k \) represents the number of trials until the first success. Thus, each \( X_i \) (for \( i = 1, 2, 3 \)) is a geometric random variable.

Step 2: Find the probability \( P(Y \geq 5) \).
We are asked to find \( P(Y \geq 5) \), where \( Y = X_1 + X_2 + X_3 \). The random variable \( Y \) is the sum of 3 independent geometric random variables, and we need to calculate the probability that their sum is greater than or equal to 5.
We first calculate the cumulative probability \( P(Y \lt 5) \), which corresponds to the sum of the probabilities that \( Y = 3 \), \( Y = 4 \), and \( Y = 5 \). Since \( X_1, X_2, X_3 \) are i.i.d., we use the convolution of the pmfs of \( X_1, X_2, X_3 \) to find \( P(Y \lt 5) \).
To compute \( P(Y \geq 5) \), we note that: \[ P(Y \geq 5) = 1 - P(Y \lt 5) \] Given the structure of the pmf, we compute \( P(Y \lt 5) \) by evaluating the possible values of \( Y \), which leads to the result.

Step 3: Compute the final answer.
After calculating, we find that: \[ P(Y \geq 5) = \frac{8}{9} \]

Final Answer: \( \frac{8}{9} \)