# Poisson Process, Probability of Rolling 2 die and getting a 7

This is what I have so far for the problem:

There are 6 different ways to roll seven with 2 die, {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. If you roll the two die n times, you have a $$(\frac{1}{6})^n$$ chance of rolling a seven. If we let $$X_t$$ be the process of rolling a die, $$X_t ~$$ Poisson$$(\lambda)$$. Let $$X_t = k$$ be the event you roll a seven.

Then, according to my professor, this is the rest of the problem. However, I don't understand how he got this distribution and simplified it.

Then, $$P(X_t = k) = \sum_{n=0}^{\infty} (\frac{1}{6})^n * \frac{e^{\lambda t}(\lambda t)^n}{n!} = e^\frac{-5\lambda t}{6}$$.

You have an error in your exponential. Its $$e^{-\lambda t}$$. Then you use conditional probabilities and denote the poisson process by $$(N_t)_{t\geq 0}$$:

$$P(X_t = k)=\sum_{n=0}^{\infty} P(X_t = k| N_t = n)P(N_t = n)=\sum_{n=0}^{\infty} \left(\dfrac{1}{6}\right)^n P(N_t = n)=\sum_{n=0}^{\infty} \left(\dfrac{1}{6}\right)^n \dfrac{e^{-\lambda t}(\lambda t)^n}{n!}$$

Now $$\left(\dfrac{1}{6}\right)^n (\lambda t )^n=\left(\dfrac{\lambda t}{6}\right)^n$$. Hence,

$$\sum_{n=0}^{\infty} \left(\dfrac{1}{6}\right)^n \dfrac{e^{-\lambda t}(\lambda t)^n}{n!}=e^{-\lambda t}e^{\frac{\lambda t}{6}} = e^{-\frac{5\lambda t}{6}}.$$