Probabilities in picking (logistics)

Question

We are in a warehouse. Inside there are picking areas constituted of picking lanes.

Every lane has the same size and the same amount of P positions where objects are stored. We have here L lanes.

Now we want to operate batch picking in these areas. A given batch needs to go to p positions of the lanes (equiproblably).

When p is given, how to determinate how many lanes are needed to go through ?

Edit (in hope to clarify) : The problem is equivalent to the following one :

• L colors
• P balls per color
• All the balls are in a bag
• We pick randomly p balls : how many different colors do we obtain ?

In meantime, I have tried to formalize the result like this (but I don't know if it is correct) :

A serie ci, where ci is the amount of colors for i picked balls.

c0 = 0

c1 = 1 = c0 + (L - c0) / L

c2 = c1 + (L - c1) / L

c3 = c2 + (L - c2) / L

...

ci+1 = ci + (L - ci) / L

So: ci+1 = 1 + ci (L - 1) / L

Answer 1

There is some vagueness in the problem statement. I'm going to interpret the question as asking for the expected number of lanes needed in order to pick a batch.

Let $$X_i = \begin{cases} 1 &\text{if lane i is chosen}\\ 0 &\text{otherwise} \end{cases}$$ for $i = 1, 2, 3, \dots , L$. Then $$\Pr(X_i=1) = 1- \left( \frac{L-1}{L} \right)^p$$ so $$E \left(\sum_{i=1}^L X_i \right) = \sum_{i=1}^L E(X_i) = L \left[ 1- \left( \frac{L-1}{L} \right)^p \right]$$ Here we have made use of the theorem that $E(X+Y) = E(X) + E(Y)$.