91Ó°ÊÓ

The expected number of objects.

Chosen by both A and B.

Substitute the A and B,

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

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

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

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

Choose $A$and $B$:

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

Multiply the value,

$=0.3×0.3$

$=0.09$

$n=10$items are total.

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

$E(Aâˆ\copyright B)=nP(Aâˆ\copyright B)$

Substitute the value,

$=10×0.09$

$=0.9$.

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

The expected number of objects.

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

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:

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

Substitute the value,

$=0.7×0.7$

$=0.49$.

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

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

Multiply the value,

$=10×0.49$

$=4.9$

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

The expected number of objects.

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

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

Simplify ,

$P\left(Aâˆ\copyright B^{\text{'}}\right)=0.3×0.7$

$=0.21$

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

$0.21+0.21=0.42$.

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

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

Simplify the value,

$=10×0.42$

$=4.2$.

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