Question

If $101$items are distributed among $10$ boxes, then at least one of the boxes must contain more than $10$ items. Use the probabilistic method to prove this result.

Short answer

Yes, there is a box containing $>10$items.

Step-by-step solution

Given information

There are $101$items that are distributed in among $10$boxes.

Explanation

Define random variables $X_i,i=1,\dots ,10$that marks the number of items in box $i$. Observe that

$E\left(X_i\right)=\frac{101}{10}$

Now, suppose that all boxes have $\le 10$items in them. Then, we have that

$101=\sum _{i}E\left(X_i\right)\le \sum _{i}10=100$

Final Answer

$101=\sum _{i}E\left(X_i\right)\le \sum _{i}10=100$

Which is a contradiction. So, there has to be a box that has$>10$ items.