Suppose on a particular day, the probability (among the entire population) of getting into a car accident is 0.04, the probability of being a texter-and-driver is 0.14, and p(car accident or being a texter-and-driver)=0.15. Find the probability a person was in a car accident given that they are a texter-and-driver. Is this higher or lower than the probability among the general population and why?

0.2143

Step-by-step explanation:

Let A be the event denote that getting into a car accident and B be the event denote being a texter-and-driver.

Thus, P(A)=0.04, P(B)=0.14.

P(A or B)=0.15.

We have to find P(A/B).

P(A/B)=P(A and B)/P(B)

P(A and B)= P(A)+P(B)-P(A or B)

P(A and B)=0.04+0.14-0.15

P(A and B)=0.03.

Thus, P(A/B)=0.03/0.14

P(A/B)=0.2143 (rounded to four decimal places)

Thus, the probability a person was in a car accident given that they are a texter-and-driver is 0.2143. This probability is higher than the probability among the general population because texting while driving is fatal.

answer:

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step-by-step explanation: