Chapter 6.4, Problem 41E

a.

To determine

1. i. Find the probability, $P\left(A_1\right)$.
2. ii. Find the probability, $P\left(A_1\cap S\right)$.
3. iii. Find the probability, $P\left(A_1|S\right)$.
4. iv. Find the probability, $P\left(\text{Not }{A}_1\right)$.
5. v. Find the probability, $P\left(S|A_1\right)$.
6. vi. Find the probability, $P\left(S|A_6\right)$.

Answer to Problem 41E

The calculated probabilities are as follows:

1. i. $P\left(A_1\right)=0.167$.
2. ii. $P\left(A_1\cap S\right)=0.068$.
3. iii. $P\left(A_1|S\right)=0.238$.
4. iv. $P\left(\text{Not }{A}_1\right)=0.833$.
5. v. $P\left(S|A_1\right)=0.41$.
6. vi. $P\left(S|A_6\right)=0.18$.

Explanation of Solution

Calculation:

The data summarize the seatbelt usage by age.

i.

Event S denotes that a randomly selected individual uses seat belt regularly, A₁ denotes the event that a randomly selected individual is in the age group “18-24”, and A₆ denotes the event that a randomly selected individual is in the age group “65 and older”.

The row and column totals are given in the following table:

Age	Does not use seat belt regularly	Uses seat belt regularly	Total
18-24	59	41	100
25-34	73	27	100
35-44	74	26	100
45-54	70	30	100
55-64	70	30	100
65 and older	82	18	100
Total	428	172	600

The probability of any event B is given below:

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

The probability that an adult in the age group 18-24 does not use seat belt is obtained by dividing the number of individuals who do not use seat belt regularly with the total number of individuals considered.

The total number of individuals who do not use seat belt in the age group 18-24 is 59.

The total number individuals considered are 600.

The probability that an adult in the age group 18-24 does not use seat belt is calculated as follows:

$\begin{array}{c}P\left(\begin{array}{l}\text{Does not use seatbelt}\\ \text{in age group 18-24}\end{array}\right)=\frac{59}{600}\\ =0.09833\end{array}$

Thus, the probability in the first cell is obtained. Similarly, the probability values in the other cells are also calculated.

Age	Does not use seat belt regularly	Uses seat belt regularly	Total
18-24	0.09833	0.06833	0.16667
25-34	0.12167	0.04500	0.16667
35-44	0.12333	0.04333	0.16667
45-54	0.11667	0.05000	0.16667
55-64	0.11667	0.05000	0.16667
65 and older	0.13667	0.0300	0.16667
Total	0.7133	0.28667	1.000

It is given that A₁ denotes the event that a randomly selected individual is in the age group 18-24.

From the above table, it is clear that probability that a randomly selected individual is in the age group 18-24 is 0.16667.

Thus, $P\left(A_1\right)=0.167$.

ii.

The required probability is $P\left(A_1\cap S\right)$.

Event S denotes that a randomly selected individual uses seat belt regularly, and A₁ denotes the event that a randomly selected individual is in the age group 18-24. From the table of probabilities, it can be observed that the probability of an individual is in the age group 18-24 and the individual uses seat belt regularly 0.06833.

That is, $P\left(A_1\cap S\right)=0.068$.

iii.

Conditional rule:

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

The required probability is $P\left(A_1|S\right)$.

$P\left(A_1|S\right)=\frac{P\left(A_1\cap S\right)}{P\left(S\right)}$

From the table of probabilities, it can be observed that the probability of an individual is in the age group 18-24, and the individual uses seat belt regularly 0.06833. Also, the probability that individual uses seat belt regularly is 0.28667.

$\begin{array}{c}P\left(A_1|S\right)=\frac{P\left(A_1\cap S\right)}{P\left(S\right)}\\ =\frac{0.06833}{0.28667}\\ =0.238\end{array}$

That is, $P\left(A_1|S\right)=0.238$.

iv.

The required probability is $P\left(\text{Not }{A}_1\right)$.

It is given that A₁ denotes the event that a randomly selected individual is in the age group 18-24.

From Part (i), it is clear that probability that a randomly selected individual is in the age group 18-24 is 0.16667. That is, $P\left(A_1\right)=0.167$.

The required probability can be obtained as the compliment of the probability that a randomly selected individual is in the age group 18-24.

The probability is calculated as follows:

$\begin{array}{c}P\left(\text{Not }{A}_1\right)=1-0.167\\ =0.833\end{array}$

Thus, $P\left(\text{Not }{A}_1\right)=0.833$.

v.

The required probability is $P\left(S|A_1\right)$.

$P\left(S|A_1\right)=\frac{P\left(S\cap A_1\right)}{P\left(A_1\right)}$

From the table of probabilities, it can be observed that the probability of an individual is in the age group 18-24 and the individual uses seat belt regularly 0.06833. Also, the probability that individual is in the age group 18-24 is 0.16667.

$\begin{array}{c}P\left(S|A_1\right)=\frac{P\left(S\cap A_1\right)}{P\left(A_1\right)}\\ =\frac{0.06833}{0.16667}\\ =0.41\end{array}$

That is, $P\left(S|A_1\right)=0.41$.

vi.

The required probability is $P\left(S|A_6\right)$.

$P\left(S|A_6\right)=\frac{P\left(S\cap A_6\right)}{P\left(A_6\right)}$

Event A₆ denotes the event that a randomly selected individual is in the age group 65 and older. From the table of probabilities, it can be observed that the probability of an individual is in the age group 65 and older and the individual uses seat belt regularly 0.03. Also, the probability that an individual is in the age group 65 and older is 0.16667.

$\begin{array}{c}P\left(S|A_6\right)=\frac{P\left(S\cap A_6\right)}{P\left(A_6\right)}\\ =\frac{0.03}{0.16667}\\ =0.18\end{array}$

That is, $P\left(S|A_6\right)=0.18$.

b.

To determine

Explain how 18-24 years and seniors differ with respect to seat belt usage using the probabilities $P\left(S|A_1\right)$ and $P\left(S|A_6\right)$ obtained in Part (a).

Explanation of Solution

From Part (a), the probability that a randomly selected individual uses seat belt regularly, given that the individual is in the age group 18-24 is 0.41. Also, the probability that randomly selected individual uses seat belt regularly, given that the individual is in the age group 65 and older is 0.18.

The probabilities state that 18-24 years are more likely to wear seat belts regularly than seniors.