<s>[INST] Finding expected value of X By rolling 6-sided fair dice, and if the outcome equals to 1 or 2, then X = 3; if it is 3 or 4, then X = 4, and if the outcome is 5 or 6, then X = 10. Find E[X].
I solved it by constructing a table of a die probability and got this:
P(X=3) = 1/36 + 2/36 = 3/36;
P(X=4) = 3/36 + 4/36 = 7/36;
P(X=5) = 5/36 + 6/36 = 11/36;
E[X] = 3*3/36 + 4*7/36 + 5*11/36 = 4.08
Is this solution correct or has any mistakes?
 [/INST] Denote $D$ to mean the value of the die roll. We observe that $P(X = 3) = P(D = 1) + P(D = 2) = \frac{1}{3}$. Similarly, $P(X=4)=P(X=10)=\frac{1}{3}$.
Thus 
$$
E[X] = \frac{1}{3}(3+4+10) = \frac{17}{3}.
$$
You could also just observe that the process is really rolling a fair "three sided" die to get the probabilities.
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