Let's break this down step by step to find both $P(X^2 = 4 \mid X > 0)$ and $E(X)$.

Step 1: Moment-Generating Function (MGF)

The moment-generating function (MGF) of the random variable $X$ is given as: $M(t) = \frac{1}{6} e^{-2t} + \frac{1}{4} e^{-t} + \frac{1}{3} e^{t} + \frac{1}{4}$ We use this to find the distribution of $X$.

The MGF typically represents a mixture of exponential distributions here because of the form of the function. We will match this form to a mixture of exponentials.

The function $M(t)$ suggests that $X$ is a mixture of four exponential distributions with the following components:

1. An exponential distribution with rate $2$ and probability $\frac{1}{6}$.
2. An exponential distribution with rate $1$ and probability $\frac{1}{4}$.
3. An exponential distribution with rate $-1$ and probability $\frac{1}{3}$.
4. A constant value 0 with probability $\frac{1}{4}$.

Thus, the random variable $X$ can take the following values with these probabilities:

• $X = 0$ with probability $\frac{1}{4}$,
• $X \sim \text{Exp}(2)$ with probability $\frac{1}{6}$,
• $X \sim \text{Exp}(1)$ with probability $\frac{1}{4}$,
• $X \sim \text{Exp}(-1)$ with probability $\frac{1}{3}$.

Step 2: $P(X^2 = 4 \mid X > 0)$

We are asked to compute $P(X^2 = 4 \mid X > 0)$. Notice that $X^2 = 4$ implies $X = 2$ (since $X > 0$).

Thus, $P(X^2 = 4 \mid X > 0) = P(X = 2 \mid X > 0)$.

This is a conditional probability, so we use the formula: $P(X = 2 \mid X > 0) = \frac{P(X = 2)}{P(X > 0)}$ Now, let’s compute each part:

1. $P(X = 2)$: From the MGF, we can infer that $X = 2$ is part of the exponential distribution with rate $1$. The probability mass function for $X$ is weighted, and based on the exponential distribution and the given probabilities, the value of $P(X = 2)$ can be obtained from the coefficients of the moment generating function. We take a weighted sum.

2. P(X>0): Would involve ensuring the weighted.