Question

Let \( X \) be a continuous random variable having the moment generating function \[ M(t) = \frac{e^t - 1}{t}, \quad t \ne 0. \] Let \( \alpha = P(48X^2 - 40X + 3>0) \) and \( \beta = P((\ln X)^2 + 2\ln X - 3>0) \). Then, the value of \( \alpha - 2\ln \beta \) is equal to:

• A. \(\frac{10}{3}\)
• B. \(\frac{19}{3}\)
• C. \(\frac{13}{3}\)
• D. \(\frac{17}{3}\)

Hint:

The MGF \( \frac{e^t - 1}{t} \) identifies the uniform distribution \( U(0,1) \); always check for valid ranges of transformed variables like \(\ln X\).

Answer 1

The Correct Option is B

Step 1: Identify the distribution from MGF.
Given \( M(t) = \frac{e^t - 1}{t} \), this is the MGF of a \( U(0,1) \) random variable, i.e., \( X \sim U(0,1) \).
Step 2: Simplify \( \alpha = P(48X^2 - 40X + 3>0) \).
Solve \( 48X^2 - 40X + 3 = 0 \): \[ X = \frac{40 \pm \sqrt{(-40)^2 - 4(48)(3)}}{2(48)} = \frac{40 \pm 32}{96} \] \[ X = \frac{1}{12}, \; \frac{3}{4} \] Since the quadratic opens upward, \( 48X^2 - 40X + 3>0 \) for \( X<\frac{1}{12} \) or \( X>\frac{3}{4} \). Thus, \[ \alpha = P(X<\tfrac{1}{12}) + P(X>\tfrac{3}{4}) = \tfrac{1}{12} + \tfrac{1}{4} = \tfrac{1}{3}. \]
Step 3: Simplify \( \beta = P((\ln X)^2 + 2\ln X - 3>0) \).
Let \( Y = \ln X \). The inequality becomes \( Y^2 + 2Y - 3>0 \Rightarrow (Y+3)(Y-1)>0 \). Hence \( Y<-3 \) or \( Y>1 \). Since \( X = e^Y \in (0,1) \), \( Y>1 \Rightarrow X>e \) is invalid, only \( Y<-3 \) holds. Thus, \( \beta = P(X<e^{-3}) = e^{-3}. \)
Step 4: Compute the expression.
\[ \alpha - 2\ln \beta = \frac{1}{3} - 2\ln(e^{-3}) = \frac{1}{3} - 2(-3) = \frac{1}{3} + 6 = \frac{19}{3}. \] Final Answer: \[ \boxed{\frac{19}{3}} \]