Question

Let \( X \) be a non-constant positive random variable such that \( E(X) = 9 \). Then which one of the following statements is true?

• A. \( E\left( \frac{1}{X+1} \right)>0.1 \) and \( P(X \geq 10) \leq 0.9 \)
• B. \( E\left( \frac{1}{X+1} \right)<0.1 \) and \( P(X \geq 10) \leq 0.9 \)
• C. \( E\left( \frac{1}{X+1} \right)>0.1 \) and \( P(X \geq 10)>0.9 \)
• D. \( E\left( \frac{1}{X+1} \right)<0.1 \) and \( P(X \geq 10)>0.9 \)

Hint:

When analyzing expectations involving transformations of a random variable, consider the nature of the transformation (e.g., reciprocals) and the behavior of the random variable around its mean to estimate the value of the expectation.

Answer 1

The Correct Option is A

Given that \( E(X) = 9 \) and \( X \) is a non-constant positive random variable, we need to analyze the expectations and probabilities provided in the options. Step 1: Understanding the expectation and probability.
The expectation \( E(X) = 9 \) tells us that on average, the random variable \( X \) takes a value around 9. From this, we can infer that for a positive random variable, values larger than 9 are possible, and thus, \( P(X \geq 10) \) is less than or equal to 1, but should reasonably be less than 0.9. Step 2: Investigating \( E\left( \frac{1}{X+1} \right) \).
The term \( E\left( \frac{1}{X+1} \right) \) involves the reciprocal of \( X+1 \), which is a decreasing function of \( X \). Given that \( E(X) = 9 \), \( \frac{1}{X+1} \) should have an average value greater than 0.1, because \( X+1 \) will mostly hover around 10, and the reciprocal of 10 is 0.1. Therefore, it is reasonable to expect that \( E\left( \frac{1}{X+1} \right)>0.1 \). Step 3: Conclusion.
From the analysis, the most plausible option is (A), where both conditions \( E\left( \frac{1}{X+1} \right)>0.1 \) and \( P(X \geq 10) \leq 0.9 \) hold true. Final Answer: \[ \boxed{\text{(A) } E\left( \frac{1}{X+1} \right)>0.1 \text{ and } P(X \geq 10) \leq 0.9} \]