Sure! Here are three distinct probability mass functions (PMFs) along with their required tasks (either deriving the Moment Generating Function (MGF) or finding the Variance using MGF):

1. Bernoulli Distribution

PMF: $P(X = 1) = p, \quad P(X = 0) = 1 - p, \quad \text{where } 0 \leq p \leq 1$

Task: Derive the MGF of $X$ and find $Var(X)$ using the MGF.

Solution:

The MGF of a random variable $X$ is defined as: $M_X(t) = \mathbb{E}[e^{tX}]$

For a Bernoulli distribution: $M_X(t) = (1-p)e^{t \cdot 0} + p e^{t \cdot 1} = 1 - p + p e^t$

First derivative of MGF: $M_X'(t) = p e^t$ Evaluate $M_X'(t)$ at $t = 0$ to get the expected value $\mathbb{E}[X]$: $\mathbb{E}[X] = M_X'(0) = p$

Second derivative of MGF: $M_X''(t) = p e^t$ Evaluate $M_X''(t)$ at $t = 0$ to get the second moment $\mathbb{E}[X^2]$: $\mathbb{E}[X^2] = M_X''(0) = p$

The variance is: [ Var(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2 = p - p^2 = p(1 - p) ]

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2. Poisson Distribution

PMF: $P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \dots, \lambda > 0$

Task: Derive the MGF of $X$ and find $Var(X)$ using the MGF.

Solution:

The MGF of a Poisson random variable $X$ is: [ M_X(t) = \mathbb{E}[e^{tX}] = \