Chapter 4.5, Problem 115E

(a)

To determine

To find: The missing probabilities in the table.

Answer to Problem 115E

Solution: The complete table is obtained as follows.

		Gender		Total
		Men	Women
Institution	4-year institution	0.2684	0.3416	0.61
	2-year institution	0.1599	0.2301	0.39
Total		0.4283	0.5717	1

Explanation of Solution

Calculation: Let $M$ and $F$ denote the gender of the students as “Male” and “Female.” Let $4Y$ denote that the student attends a 4-year institution and $2Y$ denote that the student attends a 2-year institution. The conditional probability, which is defined as the occurrence of one event provided the other event has already occurred, is used in this exercise. The probability that a male student attends a 4-year institution is denoted as $P\left(M|4Y\right)$ and the probability that a male student attends a 2-year institution is denoted as $P\left(M|2Y\right)$.

It is provided in the question that the 4-year institutions constitute 44% males; 61% of students attend the 4-year institution and the rest 2-year institution.

The probability that a student attends a 4-year institution is as follows:

$\begin{array}{c}P\left(4\text{-year institution}\right)=P\left(4Y\right)\\ =0.61\end{array}$

The probability that a male student attends a 4-year institution is as follows:

$\begin{array}{c}P\left(\text{Males|}4\text{-year instituion}\right)=P\left(M|4Y\right)\\ =44\%\\ =0.44\end{array}$

The probability that a male student attends a 2-year institution is as follows:

$\begin{array}{c}P\left(\text{Males|}2-\text{year instituion}\right)=P\left(M|2Y\right)\\ =41\%\\ =0.41\end{array}$

The provided probabilities are represented in the following table:

		Gender		Total
		Men	Women
Institution	4-year institution			0.61
	2-year institution
Total				1

The remaining or missing probabilities can be ascertained as follows.

The probability that a student attends a 4-year institution is calculated as

$\begin{array}{c}P\left(2\text{-year institution}\right)=P\left(2Y\right)\\ =1-P\left(4\text{-year institution}\right)\\ =1-P\left(4Y\right)\\ =1-0.61\end{array}$

$=0.39$

The concept of conditional probability is the probability of an event $\left(A\right)$ when the other event has already occurred $\left(B\right),$ and it is calculated as

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

The concept of conditional probability is used to ascertain the following probabilities.

The probability that a student is a male and also that he attends a 4-year institution is calculated as follows:

$\begin{array}{c}P\left(\text{Males|}4\text{-year institution}\right)=\frac{P\left(\text{Males who attend 4-year institution}\right)}{P\left(4\text{-year institution}\right)}\\ P\left(M|4Y\right)=\frac{P\left(M\cap 4Y\right)}{P\left(4Y\right)}\\ 0.44=\frac{P\left(M\cap 4Y\right)}{0.61}\\ P\left(M\cap 4Y\right)=0.44\times 0.61\end{array}$

$=0.2684$

The probability that a student is a male and also that he attends a 2-year institution is calculated as follows:

$\begin{array}{c}P\left(\text{Males|}2\text{-year institution}\right)=\frac{P\left(\text{Males who attend 2-year institution}\right)}{P\left(2\text{-year institution}\right)}\\ P\left(M|2Y\right)=\frac{P\left(M\cap 2Y\right)}{P\left(2Y\right)}\\ 0.41=\frac{P\left(M\cap 2Y\right)}{0.39}\\ P\left(M\cap 2Y\right)=0.41\times 0.39\end{array}$

$=0.1599$

The probability that the student is a female and also that she attends a 4-year institution is calculated as follows:

$\begin{array}{c}P\left(\text{Female | }4\text{-year institution}\right)=\left(P\left(4\text{-year institution}\right)-P\left(\begin{array}{l}\text{Males who attend}\\ \text{ 4-year institution}\end{array}\right)\right)\\ P\left(F\cap 4Y\right)=P\left(4Y\right)-P\left(M\cap 4Y\right)\\ =0.61-0.2684\\ =0.3416\end{array}$

The probability that the student is a female and also that she attends a 2-year institution is calculated as follows:

$\begin{array}{c}P\left(\text{Female|}2\text{-year institution}\right)=\left(P\left(2\text{-year institution}\right)-P\left(\begin{array}{l}\text{Males who attend}\\ \text{ 2-year institution}\end{array}\right)\right)\\ P\left(F\cap 2Y\right)=P\left(2Y\right)-P\left(M\cap 2Y\right)\\ =0.39-0.1599\\ =0.2301\end{array}$

Therefore, the probability that a student is a male and belongs to the 4-year institution $\left\{P\left(M\cap 4Y\right)\right\}$ is 0.2684.

The probability that a student is a male and belongs to the 2-year institution $\left\{P\left(M\cap 2Y\right)\right\}$ is 0.1599.

The probability that a student is a female and belongs to the 4-year institution $\left\{P\left(F\cap 4Y\right)\right\}$ is 0.3416.

The probability that a student is a female and belongs to the 2-year institution $\left\{P\left(F\cap 2Y\right)\right\}$ is 0.2301.

The probabilities $P\left(4Y\right)$ and $P\left(2Y\right)$ sum to 1. Also, the $P\left(M\right)$ and $P\left(W\right)$ sum to 1.

Therefore, the complete probability table is shown as below:

		Gender		Total
		Men	Women
Institution	4-year institution	0.2684	0.3416	0.61
	2-year institution	0.1599	0.2301	0.39
Total		0.4283	0.5717	1

(b)

To determine

To find: The probability that a randomly selected female student attends the 4-year institution.

Answer to Problem 115E

Solution: The required probability is 0.5975.

Explanation of Solution

Calculation: The concept of conditional probability is the probability of an event (A) when the other event has already occurred (B), and it is calculated as

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

The conditional probability rule is used to ascertain the required probability. It is calculated as follows.

The probability that the randomly selected student attends a 4-year institution provided the fact that the student is a female is calculated as follows:

$\begin{array}{c}P\left(\text{4-year institution|female}\right)=\frac{P\left(\text{Female who attend 4-year institution}\right)}{P\left(\text{Female}\right)}\\ =\frac{P\left(F\cap 4Y\right)}{P\left(F\right)}\\ =\frac{0.3416}{0.5717}\\ =0.5975\end{array}$

Hence, the required probability is 0.5975.