Finding Expectation Suppose that A and B each randomly and independently choose 4 out of 12 objects. Find the expected number of objects chosen by both A and B.
My attempt:
$$X_i = 1$$ when ith is chosen by A
$$X_i = 0$$ otherwise 
Similarly I define another indicator $Y_i$ for person B which says the exact same thing as $X_i$ but for B.
Now, $$Z=\sum(X_iY_i)$$ $i=0,\dots12$
$$\mathbb{E}(X_i)=\mathbb{E}(Y_i)= \frac{1}{12}$$
And $$\mathbb{E}[X_iY_i] = 1/12^2$$
And so the expected number chosen by both should be $$\frac{12}{12^2} = \frac{1}{12}$$
However this is not the correct answer, so can someone give me the solution and tell me where I went wrong.
 A: Let $Z_i$ take value $1$ if object $i$ is chosen by $A$ or by $B$.
Then $$Z=Z_1+\cdots+Z_{12}$$is the number of objects chosen by $A$ or by $B$.
With linearity of expectation and symmetry we find:$$\mathbb EZ=12\mathbb EZ_1=12P(Z_1=1)$$
Here $$P(Z_1=1)=$$$$P(1\text{ is chosen by }A\text{ or } 1\text{ is chosen by }B)=$$$$P(1\text{ is chosen by }A)+P(1\text{ is chosen by }B)-P(1\text{ is chosen by }A\text{ and } B)=$$$$\frac13+\frac13-\frac13\frac13=\frac59$$so the final answer is: $$12\frac59=\frac{20}3$$

edit: (I was attended on a misinterpretation of your question)
Let $U_i$  take value $1$ if object $i$ is chosen by $A$ and $B$.
Then $$U=U_1+\cdots+U_{12}$$is the number of objects chosen by $A$ and by $B$.
With linearity of expectation and symmetry we find:$$\mathbb EU=12\mathbb EU_1=12P(U_1=1)=12\frac19=\frac43$$
Observe that: $$4+4=\mathbb EZ+\mathbb EU$$as it should. 
A: Let's compute the number of items not chosen by any of them. 
The number of items not chosen by either of them would be 
$$12\left(\frac{8}{12}\right)^2= 12\left( \frac{2}{3}\right)^2$$
Hence the number of item chosen by at least one of them is 
$$12 \left( 1-\frac{4}{9}\right)= 12\left( \frac59 \right)= \frac{20}3$$
Edit:
The expected number of items that are chosen by both of them  would be 
$$12 \left( \frac13 \right)^2=\frac43$$
