Chapter 4.3, Problem 34E

a.

To determine

To obtain: Theprobability that the adoption was from Ethiopia given that it was from 2010.

Answer to Problem 34E

The probabilitythat the adoption was from Ethiopia given that it was from 2010 is 0.359.

Explanation of Solution

Given info:

The data shows that the numbers offoreign adoptions from the United States.

Calculation:

The total number of adoptions is shown in the Table (1).

	2006	2010	Total
China	6,493	3,401	9,894
Ethiopia	732	2,513	3,245
Russia	3,706	1,082	4,788
Total	10,931	6,996	17,927

Table (1)

Let event A denote that theadoptionwas from 2010 and event B denote thatthe adoption was from Ethiopia.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute ‘6,996’ for ‘Number of outcomes in A’ and 17,927 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(A\right)=\frac{6,996}{17,927}\\ =0.39\end{array}$

The formula for probability of event A and B is,

$P\left(A\text{ and }B\right)=\frac{\text{Number of outcomes in }A\text{ and }B}{\text{Total number of outcomes in the sample space}}$

Substitute ‘2,513’for ‘Number of outcomes in A and B’ and 17,927 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(A\text{ and }B\right)=\frac{2,513}{17,927}\\ =0.14\end{array}$

Conditional rule:

The formula for probability of A given B is, $P\left(B|A\right)=\frac{P\left(A\text{ and }B\right)}{P\left(A\right)}$

Substitute 0.14 for ‘ $P\left(A\text{ and }B\right)$’ and 0.39for $P\left(A\right)$.

The required probability is,

$\begin{array}{c}P\left(B|A\right)=\frac{P\left(A\text{ and }B\right)}{P\left(A\right)}\\ =\frac{0.14}{0.39}\\ =0.359\end{array}$

Thus, the probability that the adoption was from Ethiopia given that it was from 2010 is 0.359.

b.

To determine

To obtain: The probability that the adoptionwas from Russia and in 2006.

Answer to Problem 34E

The probability that the adoptionwas from Russia and in 2006is 0.074.

Explanation of Solution

Let event C denote that the adoption was from Russia and event D denote that adoption was in 2006.

The formula for probability of event C and D is,

$P\left(C\text{ and }D\right)=\frac{\text{Number of outcomes in }C\text{ and }D}{\text{Total number of outcomes in the sample space}}$

Substitute 3,706 for ‘Number of outcomes in C and D’ and 17,927 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(C\text{ and }D\right)=\frac{3,706}{17,927}\\ =0.207\end{array}$

Thus, the probability that the adoptionwas from Russiaand in 2006 is 0.207.

c.

To determine

To obtain: The probability that the adoptiondid not occur in 2006 and was not from Ethiopia.

Answer to Problem 34E

The probability that the adoption did not occur in 2006 and was not from Ethiopiais 0.25.

Explanation of Solution

Let event E denote that theadoptionwas from Ethiopiaand event D denote thatadoptionwas from 2006.

The formula for probability of event $\overline{E}$ and $\overline{D}$ is,

$P\left(\overline{E}\text{ and }\overline{D}\right)=\frac{\text{Number of outcomes in }\overline{E}\text{ and }\overline{D}}{\text{Total number of outcomes in the sample space}}$

Substitute ‘4,483 $=\left(3,401+1,082\right)$ for ‘Number of outcomes in $\overline{E}$ and $\overline{D}$’ and 17,927 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(\overline{E}\text{ and }\overline{D}\right)=\frac{4,483}{17,927}\\ =0.25\end{array}$

Thus, the probability that the adoptionwas did not occur in 2006 and was not from Ethiopiais 0.25.

d.

To determine

To obtain: The probabilitythatboth the adoptionsare from china.

Answer to Problem 34E

The probabilitythat both the adoptionsare from china is

Explanation of Solution

Given info:

There are 2 adoptionschosen at random.

Calculation:

Multiplicationrule:

If the A and B are independent, then $P\left(A\text{ and }B\right)=P\left(A\right)P\left(B\right)$.

Let event F denote that theadoptionwas from china.

The formula for probability of event F is,

$P\left(F\right)=\frac{\text{Number of outcomes in }F}{\text{Total number of outcomes in the sample space}}$

Substitute 9,894 for ‘Number of outcomes in F’ and 17,927 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(F\right)=\frac{9,894}{17,927}\\ =0.552\end{array}$

Hence, the probabilitythat theadoptionwas from chinais 0.552.

Each adoption is independent of the other.

By applying multiplication rule, the required probability is,

$\begin{array}{c}P\left(F\text{ and }F\right)=P\left(F\right)\times P\left(F\right)\\ =0.552\times 0.552\\ =0.305\end{array}$

Therefore, the probabilitythat both the adoptionsare from china is 0.305.

Interpretation:

There is a 0.305 probabilitythat both the adoptionsare from china.