Q. 7.7 - A First Course in Probability

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.  

Step-by-Step Solution

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 $E(" A$or$B ")=4.2$.

Step 1: Given Information (part a)

The expected number of objects.

Chosen by both A and B.

Step 2: Explanation (part a)

Substitute the  A and B, 

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

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

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

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

Step 3: 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, 

$=10\times 0.09$

$=0.9$.

Step 4: Final answer (part a)

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

Step 5: Given Information (Part b)

The expected number of objects.

Not chosen by either $A$ and$B$.

Step 6: 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,

$=0.7\times 0.7$

$=0.49$.

Step 7: 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,

$=10\times 0.49$

$=4.9$

Step 8: Final answer (Part b)

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

Step 9: Given Information (Part c)

The expected number of objects.

Chosen by exactly one of$A$ and $B$.

Step 10: 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,

$0.21+0.21=0.42$.

Step 11: Explanation (Part c)

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

 $E("A$ or$B")$

Simplify the value,

$=10\times 0.42$

$=4.2$.

Step 12: Final answer (Part c)

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