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?
On
The total probability is 1. Probability of rolling 1 or 2 is equal to probability of rolling 3 or 4 which is equal to probability of rolling 5 or 6. In each case it’s 1/3 (since 2/6 numbers). So EV is equal to 17/3 ie 5 and 2/3
(3*(1/3)+ 4*(1/3) + 10 *(1/3)).
Or another way of doing is is simply (1/6)*3 + (1/6)*3 + (1/6)*4 + (1/6)*4 +(1/6)*10+ (1/6)*10
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.