Exercise question on expectation The question:
A purse contains four coins each of which is either a one cent coin or a 2 cents coin. Find the expected value of a coin in that purse.
My solution:
$\dfrac{3}{2}$ 
I assigned  $1$ and $2$ as random variables whose probabilities are $\dfrac{1}{2}$.
Subsequently,
$$1\cdot\frac{1}{2} + 2\cdot \frac{1}{2} = \frac{3}{2}$$
Answer in the exercise book : 6 No explanation provided.
Could someone tell me if my answer is correct?And how is it $6$, if $6$ is the correct answer? 
 A: Either the given answer is wrong or you did not give the exact wording. I will assume the questions asked for the expected value of the coins in that purse.
Let us suppose that the coins have labels $1,2,3,4$. Let $X_i$ be the value of the $i$-th coin. You correctly found that for each $i$, we have $E(X_i)=\frac{3}{2}$.
But the total value of the coins is $X_1+X_2+X_3+X_4$. The exectation of a sum is the sum of the expectations, so we get $\frac{3}{2}+\frac{3}{2}+\frac{3}{2}+\frac{3}{2}$.
More informally, each coin has expected value $\frac{3}{2}$, so the $4$ coins together have expected value $(4)\left(\frac{3}{2}\right)$.
But with the wording a coin, your answer is perfectly correct.
A: The wording is a little wrong, The question is what is the Expected value of total coins in purse. The answer is $6$.
Total ways of having $4$ coins in a bag is $16$:
a. $6$ ways of two 1s and two 2s.
b. $4$ ways of three 1s and one of 2.
c. $4$ of three 2s and one of 1.
d. $1$ way for all 1s and similarly 1 of all 2s.
This amounts to 
$$\frac{6}{16}\cdot(2+2+1+1) + \frac{4}{16}\cdot(1+1+1+2)$$
and so on for all five terms. Summing them up using $E(x) = \Sigma(P(X_{i})\cdot X_{i})$, you get $6$.
