What is the expected number of times a 6 appears when a fair die is rolled 10 times?

Question

I think I have a working solution to this problem. Here's how I would solve it:

You look at a 6 appearing as a success and everything else as a failure. From here you can use the definition of expectation with the binomial probability function to get something like

(0 $\times$ P(exactly zero 6's)) + (1 $\times$ P(exactly one 6)) + ... + (10 $\times$ P(exactly ten 6's))

to solve the probability portions you use the binomial probability function.

Is this solution a valid way to solve the problem? Is there a better way to solve it? Because this method looks way too time consuming.

Answer 1

Assuming independence, you can do it easier:

Denote by $X_i$ the event "6 occured in roll i", and then:

$$E(X)=\sum_{i=1}^{10} E(X_i)=10E(X_1)=10*\frac16=\frac{10}6=1\frac23$$

EDIT: As noted below, you don't even need to assume independence.

Answer 2

As you suggest, use the binomial distribution. Note that the expected value ($X$ is the number of successes in $n$ trials, $p$ is the probability of success in any trial) is given by

$E(X)=np=10*\frac{1}{6}$.

The derivation of the formula for the expectation as given is not too hard and should be in your textbook.