We randomly place 200 balls independently in 100 boxes in the most natural uniform way. That is, each ball is placed independently from the rest of the balls in such a way that the probability to put it into the ith box is one-percent (1 ≤ i ≤ 100). Let X denote the number of empty boxes at the end. What is the expected value of X? I also want the numerical value.

For each boox any time a ball is placed, there are only two possible outcomes. Either the ball is put into the box, or it is not. The boxes are independent. So the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

200 balls placed in boxes:

This means that [tex]n = 200[/tex]

For each box, the ball has 1/100 probability of being put there:

This means that [tex]p = \frac{1}{100} = 0.01[/tex]

The probability of a box being empty:

This is P(X = 0).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{100,0}.(0.01)^{0}.(0.99)^{200} = 0.1340[/tex]

Let X denote the number of empty boxes at the end. What is the expected value of X?

Each box has a 0.1340 probability of being empty at the end.

There are 100 boxes.

So

0.1340*100 = 13.4

The expected value of X is 13.4.