1. ## Expectation Problem

Hey guys, homework problem on expectation. This is what i have so far.

Suppose a fair die is rolled ten times. Find numerical values for the expectations of:

1) the number of faces which fail to appear in the ten rolls

Here's what i got:

$E(x) = \sum x P(X = x) = 1 P(X = 1) + 2 P(X = 2) + ... + 5 P(X = 5)$

So the probability that x faces dont is ... this is the part im not sure about.
$P(X = x) = ?$

I know, for example, that $P(X = 5)$ would have to be (1/6)^10
After this i am unsure how to proceed. $P(X = 1)$ seems like it would NOT be $\left(\begin{array}{cc}10\\5\end{array}\right)(5/6)^{10}$ ... I cant figure out how to find a general formula for this.

2. ## Response

I may be missing something, but this is how I see it.

There are 6 possible faces that may not show, and for each one, the probability is going to be the same (5/6)^10. So for an expected value, I believe it's simply:

6 * (5/6)^10 = .96903

Intuitively, this seems like a reasonable answer.

3. But does that apply when *multiple* faces don't show in ten rolls? Thats what i'm looking for.

From what i understand, what you've described is the probability a certain face wont show, then multiplied by 6 for each case.

4. Juicy's answer is correct. Let X = the number of faces not rolled, and let X_i = 1 if face i isn't rolled, 0 otherwise. Then X = X_1 + ... + X_6, and by linearity of the expectation E(X) = E(X_1) + ... + E(X_6) = 6*(5/6)^10.