Step 1: Rewrite the expression.
The given inequality \( X^3 - 2X^2 - X + 2 > 0 \) involves a cubic function of \( X \), which we solve numerically or use standard normal distribution tables to evaluate.
Step 2: Numerical approximation.
By analyzing the roots of the cubic function and using the properties of the standard normal distribution, we find that: \[ P(X^3 - 2X^2 - X + 2 > 0) = 2\Phi(1) - \Phi(2). \]
Step 3: Conclusion.
The correct answer is (C) \( 2\Phi(1) - \Phi(2) \).
Let $X$ and $Y$ be independent random variables with respective moment generating functions $M_X(t) = \dfrac{(8 + e^t)^2}{81}$ and $M_Y(t) = \dfrac{(1 + 3e^t)^3}{64}$, $-\infty < t < \infty$. Then $P(X + Y = 1)$ equals .............. (round off to two decimal places).