Question

a survey was taken of automobile preferences for different age groups. they were asked whether they would prefer owning a car or a truck. find each probability as a fraction.

	truck	car	total
36 - 55	30	160	190
total	50	200	250

1. $p(car)=$
2. $p(truck|36 - 55)=$
3. $p(16 - 35$ and $car)=$

Explanation:

Step1: Recall probability formula

The probability formula is $P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.

Step2: Calculate $P(\text{Car})$

The number of people who prefer cars is 200 and the total number of people surveyed is 250. So $P(\text{Car})=\frac{200}{250}=\frac{4}{5}$.

Step3: Calculate $P(\text{Truck}|\text{36 - 55})$

The formula for conditional - probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$. In the context of the table, the number of people in the 36 - 55 age group who prefer trucks is 30, and the total number of people in the 36 - 55 age group is 190. So $P(\text{Truck}|\text{36 - 55})=\frac{30}{190}=\frac{3}{19}$.

Step4: Calculate $P(\text{16 - 35 and car})$

The number of people in the 16 - 35 age group who prefer cars is 40, and the total number of people surveyed is 250. So $P(\text{16 - 35 and car})=\frac{40}{250}=\frac{4}{25}$.

Answer:

1. $\frac{4}{5}$
2. $\frac{3}{19}$
3. $\frac{4}{25}$