I want to verify my answer to the following question:
The number of visitors in a store in one hour distributes $X\sim Poi(\lambda=3)$ each visitor purchase one item with a probability of 0.5 and two items with the same probability.
What is the prob. that exactly K items were purchased in two hours?
This is my solution:
first we define some variables which will assist us:
1) $X$ – number of customers entered the store in two hours.
2) $Y$ – number of customers who purchased 2 items.
3) $Z$ – number of items purchased in 2 hours.
Basically, we want to find $Pr(Z=k)$
$\displaystyle Pr(Z=k) = \sum_{j=\lceil k/2 \rceil}^kP(X=j)\cdot P(Y=k-j) = \sum_{j=\lceil k/2 \rceil}^k \frac{Exp(-3 \cdot 2)\cdot 3^j}{j!} \cdot \binom{j}{k-j}\cdot 0.5^j$
Explanation:
we know that if we want $k$ items we must have a least $\lceil k/2 \rceil$ visitors in the store (lower bound) and $k$ visitors (each of them purchasing one item) as upper bound, cause each customer buys at least one item.
Next, we have a deterministic number of customers purchasing 2 items in order to end up with $k$ items, but we have multiple combinations for that hence using the Binomial formula. We define "success" as a customer buying 2 items.
Is this solution correct? Is there a more intuitive solution to this problem?
Please share your thoughts 🙂
Best Answer
If $X$ denotes the number of customers entering the store in $2$ hours then $X\sim\mathsf{Poisson}(\lambda=6)$.
Splitting up $X=X_1+X_2$ where $X_i$ stands for the number of customers that buy $i$ items it can be noticed that $X_1,X_2$ are iid and $\sim\mathsf{Poisson}(\lambda=3)$.
In this situation to be found is expression for:$$P(X_1+2X_2=k)$$
So something like: $$\sum_{j=0}^{\lfloor k/2\rfloor} P(X_2=j)P(X_1=k-2j)=e^{-6}\sum_{j=0}^{\lfloor k/2\rfloor}\frac{3^{k-j}}{j!\left(k-2j\right)!}$$