2 Poisson distributions used to work out customers who visit in afternoon?

Question

A shop has a number of customers visiting in the morning that is Poisson distributed with parameter 2, and a number of afternoon visitors that is Poisson distributed with parameter 4.

i) What is the probability that all visits occur in the afternoon?

ii) What is the expected number of visitors in a day?

iii) What is the probability of 4 visitors in a day? Hint: You may use the following formula to simplify the expression that you obtain:

$\frac{1}{4!} (x+y)^4=\frac{x^{4}}{4!}+\frac{x^{3}y}{3!}+\frac{x^{2}y^{2}}{2!2!}+\frac{xy^{3}}{3!}+\frac{x^{4}}{4!}$

iv) Given that there are 4 visitors in a day, what is the probability that there are 2 visitors in the morning and 2 visitors in the afternoon?

So far I have tried...

i) If all occur in afternoon. Then none must be come in the morning??

$e^{-2} \approx 0.135$

ii) Using the properties of expectation: 4 + 2 = 6

iii) Not sure

iv) $\frac{e^{-2}\cdot 2^{2}}{2!} \cdot \frac{e^{-4}\cdot 4^{2}}{2!} = 2e^{-2} \cdot 8e^{-4}=16e^{-6} \approx 0.03966$

I've only just started learning probability a few weeks ago. Any help is appreciated. My lecturer has not provided me with answers.

Answer 1

(i): you should multiply by the probability that there are no visitors in the afternoon. (ii) and (iv) are correct.

For (iii): it is not difficult to prove that the sum of two independent Poisson distributed random variables with means $\lambda$ and $\mu$ is Poisson distributed with mean $\lambda+\mu$. In case you can't use that, just write out the probability: $$\begin{align*}\mathbf{P}(\text{4 visitors in a day})&=\sum_{i=0}^4 \mathbf{P}(i\text{ visitors in the morning})\cdot \mathbf{P}(4-i\text{ visitors in afternoon}) \\&=\sum_{i=0}^4 \frac{e^{-2}2^i}{i!}\frac{e^{-4}4^{4-i}}{(4-i)!}=\cdots\end{align*}$$