Stock and poisson distribution

Question

A shopkeeper finds that the number of orders for an electrical good averages 2 per week. At the start of the trading week, i.e. on a Monday, the shopkeeper has 5 such items in stock. Orders follow a Poisson distribuiton. If and when his stock level is down to two items during the week, he orders another four items. What are the chances that he will order another four items?

In other words, I'm asked on what the chances that the stock level is down to two items are. For this to happen, the shopkeeper needs to sell 3 items. However, if I understand the probelem correctly, he may sell 3 items on Monday and order another four, or may sell 0 items on Monday, 1 item on Tuesday, 2 items on Wednesday and then order another four. I don't know how to approach that problem. The answer in the book is 0.3233.

Answer 1

The number of orders is distributed according to the Poisson distribution with mean $\lambda = 2$. Since

$$ P_k = e^{-\lambda} \frac{\lambda^k}{k!} $$

and the shopkeeper will order four items if he sells at least three items, we want

$$ P_3+P_4+P_5+\ldots = 1-(P_0+P_1+P_2) $$

where we take advantage of the fact that $\sum_{k=0}^\infty P_k = 1$.

Answer 2

Use the formula poisson.dist(x,mu,conditional) in excel, in place of x put 2 as stock level is down to 2 items, average(mu) is 2, conditional is TRUE, so value will come 0.676676416 that shows 67% that his orders will be <=to 2. Therefore 1-0.676676416=0.323323584,i.e 32% prob that he will order another four items