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

Hint:

For sums of i.i.d. geometric random variables, use the CDF and calculate the complementary probability for the desired range.

Answer 1

The Correct Option is B

Step 1: Understanding the distribution.
The given probability mass function corresponds to a geometric distribution with parameter \( \frac{1}{3} \). This means that each \( X_i \) is the number of trials before the first success in a series of independent Bernoulli trials, where the probability of success in each trial is \( \frac{1}{3} \).
Step 2: Calculating the probability.
We need to find \( P(Y \geq 5) \), where \( Y = X_1 + X_2 + X_3 \). Since \( X_1, X_2, X_3 \) are independent, the sum \( Y \) is the sum of three independent geometric random variables. We can use the cumulative distribution function (CDF) of the geometric distribution to compute the probability. First, calculate the probability of \( Y<5 \). This is the same as the probability that all three variables \( X_1, X_2, X_3 \) are less than 5, which can be found by multiplying the individual probabilities for each variable. The probability that one random variable \( X \) is less than 5 is the cumulative probability from the geometric distribution. \[ P(X \geq 5) = \left( \frac{2}{3} \right)^4 = \frac{16}{81} \] So for \( Y = X_1 + X_2 + X_3 \), we compute the complementary probability: \[ P(Y \geq 5) = 1 - P(Y<5) = 1 - \left( \frac{8}{9} \right) \] Thus, the correct answer is \( \frac{8}{9} \), which corresponds to option (B).