Question

Let $X$ and $Y$ be two independent random variables with the cumulative distribution functions \[ F_X(x) = 1 - \left(\frac{3}{4}\right)^x, \quad x = 1,2,3,\ldots \] \[ F_Y(y) = 1 - \left(\frac{2}{3}\right)^y, \quad y = 1,2,3,\ldots \] respectively. Let $Z = \min\{X, Y\}$. Then, the probability $P(Z \ge 6)$ is

• A. $\dfrac{1}{64}$
• B. $\dfrac{1}{32}$
• C. $\dfrac{63}{64}$
• D. $\dfrac{31}{32}$

Hint:

For $\min(X,Y)$ of independent random variables, $P(Z \ge k) = P(X \ge k)P(Y \ge k)$. Use CDF complements carefully.

Answer 1

The Correct Option is B

Step 1: Probability that $Z \ge 6$.
Since $Z = \min(X, Y)$, \[ P(Z \ge 6) = P(X \ge 6, \ Y \ge 6) = P(X \ge 6)P(Y \ge 6), \] using independence.
Step 2: Find $P(X \ge 6)$ and $P(Y \ge 6)$.
For a discrete random variable, \[ P(X \ge 6) = 1 - F_X(5) = \left(\frac{3}{4}\right)^5 = \frac{243}{1024}, \] \[ P(Y \ge 6) = 1 - F_Y(5) = \left(\frac{2}{3}\right)^5 = \frac{32}{243}. \]
Step 3: Multiply to get $P(Z \ge 6)$.
\[ P(Z \ge 6) = \frac{243}{1024} \times \frac{32}{243} = \frac{32}{1024} = \frac{1}{32}. \] Wait — check options — it seems off by a power. If $Z \ge 6$ means both variables ≥ 6, and the distributions start from 1, indeed: \[ P(Z \ge 6) = \left(\frac{3}{4}\right)^5 \times \left(\frac{2}{3}\right)^5 = \left(\frac{1}{2}\right)^5 = \frac{1}{32}. \] Hence the correct answer is $\boxed{\frac{1}{32}}$, which corresponds to **(B)**.
Step 4: Conclusion.
$P(Z \ge 6) = \dfrac{1}{32}$.