Question refering to dice probability from a test.

Question

There are 4 dice in the following colors: blue, yellow, black and red. They're rolled at the same time. Consider $X_1$ to be the quantity of dice in which the face value scored was one. Let $X_2, ... , X_4$ be defined in the same manner. Let Y = $X_1X_2X_3X_4$. What values Y can get? Get $E(Y)$

Answer 1

We note that we define $E(X) = \sum\limits_{s \in S} p(s) \cdot X(s)$, where $s$ is an event in our sample space $S$, $p(s)$ is the probability of the event occurring, and $X(s)$ is the value of the event.

To solve the problem, we'll use linearity of expectation.

$ E(Y) = E(X_1X_2X_3X_4)$. Note that they're all independent. Then we use the fact that if $A,B$ independent, then $E(AB)=E(A)E(B)$.

So $E(Y) = E(X_1)E(X_2)E(X_3)E(X_4)$.

We note that for any die, the probability of rolling any integer $n \in \{1,2,\ldots,6\}$ is $\frac16$. As we have four rolls, the probability of rolling $n$ is $4 \cdot \frac16 = \frac46$. You can verify that this is equal to $E(X_1)$. As the probability is the same for all other $X_a$, the expected value will be the same as well. So $E(Y) = \left(\frac46\right)^4 = \frac{16}{81}$.