Suppose that $8$ white balls and $2$ black balls will be randomly ordered, from left to right (with all permutations of the $10$ balls equally likely), what is the expected value of the number of balls that will be between the two black balls ?

I gave it a shot and this is where I am stuck.

If $E$ denotes the event that at least one one white ball in between the two black balls, then

$P(E) = 1- P(E^c)$

$P(E) = 1- \frac{(2) (9!)}{10!}$

which gives $P(E)$ to be equal to $0.8$

I am not sure how to define the Indicator function with this information, that would lead me to answer the question.


1 Answer 1


Asociate an indicator variable ,$X_i$, to each white ball such that they take value $1$ if it lies between the two black balls. Then with this formulation, the number of balls between the two black balls will be equal to the sum of $X_i$.

Hence, to solve the problem, we just have to consider the probability that a white ball is between the two black balls. For each white ball, it has $3$ equally likely option. It can be on the left of both black balls, on the right of both black balls, or between the two black balls.

Hence, $$E\left[\sum_{i=1}^8X_i\right]=\sum_{i=1}^8P(X_i=1)=\frac83$$


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