Question

Let X₁ and X₂ be two independent and identically distributed discrete random variables having the probability mass function
\(f(x) = \begin{cases} (\frac{1}{2})^x, & x=1,2,3,... \\   0, & \text{otherwise. }  \end{cases}\)
Then P(min{X₁ , X₂} ≥ 5) equals

• A. \(\frac{1}{256}\)
• B. \(\frac{1}{512}\)
• C. \(\frac{1}{64}\)
• D. \(\frac{9}{256}\)

Answer 1

The Correct Option is A

To find the probability that the minimum of the two independent and identically distributed random variables \(X_1\) and \(X_2\) is at least 5, we need to consider the given probability mass function (PMF) of each random variable:

\[ f(x) = \begin{cases} \left(\frac{1}{2}\right)^x, & x=1,2,3,... \\ 0, & \text{otherwise}. \end{cases} \]

We are asked to find \( P(\min(X_1, X_2) \geq 5) \), which means both \(X_1\) and \(X_2\) must be at least 5.

The first step is to find \(P(X_1 \geq 5)\):

\[ P(X_1 \geq 5) = 1 - P(X_1 < 5) \]

Which can be expanded as:

\[ P(X_1 < 5) = P(X_1 = 1) + P(X_1 = 2) + P(X_1 = 3) + P(X_1 = 4) \]

Calculating these probabilities using the PMF:

\[ \begin{align*} P(X_1 = 1) &= \left(\frac{1}{2}\right)^1 = \frac{1}{2}, \\ P(X_1 = 2) &= \left(\frac{1}{2}\right)^2 = \frac{1}{4}, \\ P(X_1 = 3) &= \left(\frac{1}{2}\right)^3 = \frac{1}{8}, \\ P(X_1 = 4) &= \left(\frac{1}{2}\right)^4 = \frac{1}{16}. \end{align*} \]

Now, add these probabilities:

\[ P(X_1 < 5) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{15}{16} \]

So,

\[ P(X_1 \geq 5) = 1 - \frac{15}{16} = \frac{1}{16} \]

Since \(X_2\) has the same distribution, \(P(X_2 \geq 5) = \frac{1}{16}\) as well.

Thus, the probability that both \(X_1\) and \(X_2\) are greater than or equal to 5 is given by:

\[ P(\min(X_1, X_2) \geq 5) = P(X_1 \geq 5) \times P(X_2 \geq 5) = \frac{1}{16} \times \frac{1}{16} = \frac{1}{256} \]

The correct answer is thus:

\(\frac{1}{256}\)