# expectation in coin draw problem

I've been stuck on this question for a while. Two coins are selected at random from $3$ pennies, $2$ nickels, and $1$ dime with no replacement, and $X$ is the sum of the two coins. What is the probability function and the expected value?

My solution: For probability function for $X$,

$$f(x)=\begin{cases} \frac36\cdot\frac25,&\text{for }X=2\text{ cents}\\\\ \frac26\cdot\frac25,&\text{for }X=6\text{ cents}\\\\ \frac36\cdot\frac15,&\text{for }X=11\text{ cents}\\\\ \frac36\cdot\frac15,&\text{for }X=10\text{ cents}\\\\ \frac16\cdot\frac35,&\text{for }X=15\text{ cents}\;. \end{cases}$$

I can not seem to get a reasonable solution for $\mathrm{E}X$, because $\mathrm{E}X$ is a mean value where each value that $X$ takes on is either $2, 6, 10, 11$, or $15$ cents. I can't spend any more time on this, it's driving me nuts. Thanks!

-
 Your $f(x)$ can’t be right: the probabilities sum to only $19/30$. – Brian M. Scott Apr 30 '12 at 2:35

You are thinking in the right direction, but if you check, your probabilities don't add to $1$. For 11 cents, $\frac 36 \cdot \frac 15$ is the chance you take a penny and then a dime, but you can also take the dime first, so it needs to be doubled. For 15, you need a dime and a nickel, but in either order, so $2\cdot \frac 16 \cdot \frac 25$. 10 cents is wrong differently. Then EX is just $\sum P(X)X$, so $2P(2)+6P(6)+\ldots$

-
 Got it.. Thanks! – jay Apr 30 '12 at 2:39

The following solution is not the one you are expected to produce at this stage. I am mentioning it in order to provide some information that may be useful later. Assume we are picking the coins one at a time.

Let $X_1$ be the value of your first pick, and let $X_2$ be the value of your second pick. Then $X=X_1+X_2$. By the linearity of expectation, we have $$E(X)=E(X_1+X_2)=E(X_1)+E(X_2).$$ Note that $X_1$ and $X_2$ are not independent, but that doesn't matter.

We have $$E(X_1)=1\cdot \frac{3}{6}+5\cdot\frac{2}{6}+10\cdot\frac{1}{6}=\frac{23}{6}.$$ Exactly the same calculation shows that $E(X_2)=\dfrac{23}{6}$. (The probabilities for the various coins are the same for the second pick as for the first.)

It follows that $E(X)=\dfrac{46}{6}$.

-