Finding Expectation

Question

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.

Answer 1

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

$$12 \left( \frac13 \right)^2=\frac43$$

Answer 2

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.