Chapter 4.3, Problem 32E

Doctor Specialties Below are listed the numbers of doctors in various specialties by gender.

	Pathology	Pediatrics	Psychiatry
Male	12,575	33,020	27,803
Female	5,604	33,351	12,292

Choose one doctor at random.

a. Find P (male|pediatrician).

b. Find P (pathologist |female).

c. Are the characteristics “female” and “pathologist” independent? Explain.

Source: World Almanac.

To determine

To obtain: The probability that the doctor was male given that the doctor was pediatrician.

Answer to Problem 32E

The probability that the doctor was male given that the doctor was pediatrician is 0.498.

Explanation of Solution

Given info:

The data shows that the the numbers of doctors in various specialties by gender.

Calculation:

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

	Pathology	Pediatrics	Psychiatry	Total
Male	12,575	33,020	27,803	73,398
Female	5,604	33,351	12,292	51,247
Total	18,179	66,371	40,095	124,645

Table (1)

Let event A denote that the doctor was pediatrician and event B denote that the doctor was male.

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 66,371 for ‘Number of outcomes in A’ and 124,645 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(A\right)=\frac{66,371}{124,645}\\ =0.532\end{array}$

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 33,020 for ‘Number of outcomes in A and B’ and 124,645 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(A\text{ and }B\right)=\frac{33,020}{124,645}\\ =0.265\end{array}$

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 0.265 for ‘ $P\left(A\text{ and }B\right)$’ and 0.532 for $P\left(A\right)$.

The required probability is,

$\begin{array}{c}P\left(B|A\right)=\frac{P\left(A\text{ and }B\right)}{P\left(A\right)}\\ =\frac{0.265}{0.532}\\ =0.498\end{array}$

Thus, the probability that the doctor was male given that the doctor was pediatrician is 0.498.

To determine

To obtain: The probability that the doctor was pathologist given that the doctor was female.

Answer to Problem 32E

The probability that the doctor was pathologist given that the doctor was female is 0.109.

Explanation of Solution

Calculation:

Let event C denote that the doctor was female and event D denote that the doctor was pathologist.

The probability of event C is,

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

Substitute 51,247 for ‘Number of outcomes in C’ and 124,645 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(C\right)=\frac{51,247}{124,645}\\ =0.411\end{array}$

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 5,604 for ‘Number of outcomes in C and D’ and 124,645 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(C\text{ and }D\right)=\frac{5,604}{124,645}\\ =0.045\end{array}$

Applying the conditional probability, then the required probability is,

$\begin{array}{c}P\left(D|C\right)=\frac{P\left(C\text{ and }D\right)}{P\left(C\right)}\\ =\frac{0.045}{0.411}\\ =0.109\end{array}$

Thus, the probability that the doctor was pathologist given that the doctor was female is 0.109.

To determine

To explain: Whether the characteristics “female” and “pathologist” independent or not.

Answer to Problem 32E

The characteristics “female” and “pathologist” are not independent because $0.060\ne 0.045$

Explanation of Solution

Calculation:

Here, the event C denotes that the doctor was female and event D denotes that the doctor was pathologist.

From part (b), it is observed that the probability that the doctor was female is 0.411 and $P\left(C\text{ and }D\right)$ is 0.045.

The probability of event D is,

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

Substitute 18,179 for ‘Number of outcomes in D’ and 124,645 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(D\right)=\frac{18,179}{124,645}\\ =0.146\end{array}$

Independent events:

If $P\left(C\text{ and }D\right)=P\left(C\right)\times P\left(D\right)$ then the events C and D are independent.

Substitute 0.411for $P\left(C\right)$, 0.146 for $P\left(D\right)$ in $P\left(C\right)\times P\left(D\right)$

$\begin{array}{c}P\left(C\right)\times P\left(D\right)=0.411\times 0.146\\ =0.060\end{array}$

Here, $P\left(C\right)\times P\left(D\right)$ is not equal to $P\left(C\text{ and }D\right)$. That is, $0.060\ne 0.045$. Thus, the events C and D are not independent