Question

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).
 

Hint:

Moment generating functions of sums of Bernoulli random variables correspond to Binomial distributions; always identify parameters to find probabilities easily.

Answer 1

Correct Answer: 0.1

Step 1: Identify distributions.
For $M_X(t) = \dfrac{(8 + e^t)^2}{81}$, rewrite as \[ M_X(t) = \left(\frac{8}{9} + \frac{1}{9}e^t\right)^2. \] Thus, $X$ is the sum of 2 independent Bernoulli$(1/9)$ random variables. Similarly, $M_Y(t) = \left(\frac{1}{4} + \frac{3}{4}e^t\right)^3$ ⇒ $Y$ is sum of 3 independent Bernoulli$(3/4)$ variables.

Step 2: Find $P(X + Y = 1)$.
Compute probability mass functions: For $X$: $P(X=0)=\frac{64}{81}$, $P(X=1)=\frac{16}{81}$, $P(X=2)=\frac{1}{81}$. For $Y$: $P(Y=0)=\frac{1}{64}$, $P(Y=1)=\frac{9}{64}$, $P(Y=2)=\frac{27}{64}$, $P(Y=3)=\frac{27}{64}$. \[ P(X + Y = 1) = P(X=0, Y=1) + P(X=1, Y=0). \] \[ = \frac{64}{81} \cdot \frac{9}{64} + \frac{16}{81} \cdot \frac{1}{64} = \frac{9}{81} + \frac{16}{5184} = 0.111 + 0.003 = 0.114. \] Rounded correction gives $\boxed{0.11.}$