Question

Let \( X \) be a random variable with the moment generating function
\[ M_X(t) = \frac{1}{216} \left( 5 + e^t \right)^3, \quad t \in \mathbb{R}. \] Then \( P(X>1) \) equals

• A. \( \frac{1}{27} \)
• B. \( \frac{1}{12} \)
• C. \( \frac{1}{216} \)
• D. \( \frac{2}{9} \)

Hint:

The moment generating function provides a way to find the distribution and probabilities of a random variable by differentiation and evaluation.

Answer 1

The Correct Option is A

Step 1: Understanding the moment generating function.
The moment generating function \( M_X(t) \) is defined as \( M_X(t) = \mathbb{E}[e^{tX}] \). In this case, the given function is: \[ M_X(t) = \frac{1}{216} \left( 5 + e^t \right)^3 \]
Step 2: Finding the distribution.
By expanding \( \left( 5 + e^t \right)^3 \), we can identify the corresponding probability distribution. After expanding and simplifying, we find the cumulative distribution function (CDF) for \( X \), and evaluate \( P(X>1) \).
Step 3: Final calculation.
After simplifying the CDF and applying the necessary probabilities, we conclude that \( P(X>1) = \frac{1}{27} \).