Poisson

Probability Level 3

Let $X$ be a random variable with a Poisson distribution of parameter $\lambda = 2$ .
Let $Y$ be a random variable with a Poisson distribution of parameter $\lambda = 1$ .

If $X$ and $Y$ are independent random variables, then $P(X + Y = 4)$ can be written as $\dfrac{A}{B \cdot e^3}$ with $A, B$ coprime positive integers. Enter $A + B$

The answer is 35.

2 solutions

Nathan Zhao
Apr 5, 2021

Sum of two poisson variables is poisson with parameter lambda1+lambda2. Then we can just compute 3^4/(4!)

Matthew Wessler
Apr 6, 2020

Since the two variables are independent, we can find the mean of x + y by finding the MGF of (x+y). This yields exponential(3(e^t-1)). From here, we can identify the lambda of X+Y as 3 based off the MGF of a Poisson Distribtution. Finally, we can apply the formula for the probability mass function or PMF/PDF of a Poisson D. And this gives 27/(8*e^3) when fully simplified.

Poisson

The answer is 35.

2 solutions

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