Question

Suppose that A and B each randomly and independently choose$3$of$10$objects. Find the expected number of objects

a. Chosen by both A and B;

b. Not chosen by either A or B;

c. Chosen by exactly one of A and B.

Short answer

The expected number of objects value are,

a. Expected numbers of items chosen by both $A$and $B$value are$E(A\cap B)=0.9$

b. Expected numbers of items not chosen by both $A$or $B$value are $\mathrm{E}\left(A^{\text{'}}\cap B^{\text{'}}\right)=4.9$

c. Expected numbers of items chosen by either value are role="math" localid="1647253518420" $E("A$or$B")=4.2$.

Step-by-step solution

Given Information (part a)

The expected number of objects.

Chosen by both A and B.

Explanation (part a)

Substitute the A and B,

$P\left(A\right)=\frac{3}{10}=0.3$

$P\left(B\right)=\frac{3}{10}=0.3$

$P\left(A^{\text{'}}\right)=1-P\left(A\right)=1-0.3=0.7$

$P\left(B^{\text{'}}\right)=1-P\left(B\right)=1-0.3=0.7$.

Explanation (part a)

Choose $A$and $B$:

$\mathrm{P}(A\cap B)=P\left(A\right)P\left(B\right)$

Multiply the value,

$=0.3\times 0.3$

$=0.09$

$n=10$items are total.

Expected numbers of items chosen by both $A$and$B$is:

$E(A\cap B)=nP(A\cap B)$

Substitute the value,

$=10\times 0.09$

$=0.9$.

Final answer (part a)

Expected numbers of items chosen by both$A$and$B$$E(A\cap B)=0.9$.

Given Information (Part b)

The expected number of objects.

Not chosen by either $A$and$B$.

Explanation (Part b) 

Substitute the not chosen by either $A$and$B$,

$P\left(A^{\text{'}}\right)=0.7$- not choosing an item of $A$

$P\left(B^{\text{'}}\right)=0.7$- not choosing an item of$B$

The probability that neither $A$and $B$choses an item is:

$\mathrm{P}\left(A^{\text{'}}\cap B^{\text{'}}\right)=P\left(A^{\text{'}}\right)\cap P\left(B^{\text{'}}\right)$

Substitute the value,

$=0.7\times 0.7$

$=0.49$.

Explanation (Part b)

The expected number of items not chosen by $A$or$B$is:

$\mathrm{E}\left(A^{\text{'}}\cap B^{\text{'}}\right)=\mathrm{nP}\left(A^{\text{'}}\cap B^{\text{'}}\right)$

Multiply the value,

$=10\times 0.49$

$=4.9$

Final answer (Part b) 

Expected numbers of items not chosen by both$A$and$B$value are$E\left(A^{\text{'}}\cap B^{\text{'}}\right)=4.9$.

Given Information (Part c)

The expected number of objects.

Chosen by exactly one of$A$and$B$.

Explanation (Part c)

We choose $A$or do not choose $B$,

Simplify ,

$P\left(A\cap B^{\text{'}}\right)=0.3\times 0.7$

$=0.21$

Chooses an item$A$or$B$is,

$0.21+0.21=0.42$.

Explanation (Part c)

Expected numbers of items chosen by either$A$or$B$is:

$E("A$or$B")$

Simplify the value,

$=10\times 0.42$

$=4.2$.

Final answer (Part c)

Expected numbers of items chosen by either value are $E("A$or$B")=4.2$