Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a random set $\{a,b,c\}$ and a second set $\{e,d\}$ I draw one number first number and one from the second

Letting $X_1$ denote the first number and $X_2$ the second number find, $E(X_1)$ and $E(X_2)$ and $E(X_1+X_2)$

Please please help, it might be obvious but I just can't work it out!

share|improve this question
By uniform distribution do you mean that probability to pick a number from a set is equal to every number? i.e. $P(a)=P(b)=P(c)=1/3$? –  Tomas Dec 11 '12 at 20:47
Tomas - yes, I think he is referring to the Discrete uniform distribution (as opposed to the continuous one, that some of us are more used to). –  Conan Wong Dec 11 '12 at 20:48

2 Answers 2

$E[X_1]$ and $E[X_2]$ are simply the averages of all the elements in each set, respectively. $E[X_1+X_2]=E[X_1]+E[X_2]$.

share|improve this answer

As Tomas points out, $P(a) = P(b) = P(c) = \frac{1}{3}$. Thus,

$$E(X_1) = aP(a) + bP(b) + cP(c)$$


You can apply the same method to work out $E(X_2)$.

As Vincent points out, the linearity of Expected Value over these uniform distributions gives

$$E(X_1+X_2) = E(X_1) + E(X_2)$$

[Alternatively, you could go through all the possibilities of $X_1+X_2$, i.e. the set {$a+d, a+e, b+d,b+e,c+d,c+e$}. There are six elements in this set and the probability of getting each is $\frac{1}{6}$, because the distribution is uniform. And you could compute $E(X_1+X_2)$ in a similar fashion as above.]

share|improve this answer
this is what I first thought but I figured it couldn't be that simple, thanks a lot :) –  Justin Dec 11 '12 at 20:59
You're welcome, Justin :-) –  Conan Wong Dec 11 '12 at 21:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.