Question

a city planner surveyed 154 residents. for each, the planner recorded whether the resident owns a car and whether the resident lives alone. the results are summarized in the table below.

	alone	not alone
no car	32	95

suppose a resident from the survey is chosen at random.
answer each part. do not round intermediate computations, and round your answers to the nearest hundredth.

(a) what is the probability that the resident does not own a car and lives alone?

(b) what is the probability that the resident does not own a car or lives alone?

Explanation:

Step1: Find total number of residents

The total number of residents surveyed is $9 + 18+32 + 95=154$.

Step2: Calculate probability for part (a)

The number of residents who do not own a car and live alone is 32. The probability $P(\text{no car and alone})$ is the number of favorable outcomes divided by the total number of outcomes. So $P(\text{no car and alone})=\frac{32}{154}\approx 0.21$.

Step3: Calculate probability for part (b)

Let $A$ be the event that the resident does not own a car and $B$ be the event that the resident lives alone.
The number of residents who do not own a car is $32 + 95 = 127$, so $P(A)=\frac{127}{154}$.
The number of residents who live alone is $9+32 = 41$, so $P(B)=\frac{41}{154}$.
The number of residents who do not own a car and live alone is 32, so $P(A\cap B)=\frac{32}{154}$.
By the addition - rule of probability $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
$P(A\cup B)=\frac{127 + 41-32}{154}=\frac{136}{154}\approx 0.88$.

Answer:

(a) $0.21$
(b) $0.88$