Question

The police department is keeping track of distracted drivers and accidents. They have found that if a driver is distracted, the driver has a 30% chance of being an accident. If the driver is not distracted, the driver has a 2% chance of being in an accident. The probability of a driver being distracted is 10%

Answer 1

what we are asked to calculate is not included in your question. Here they are below:

a. what is the probability that a driver will be in an accident?

b. what is the Probability that a driver that was in an accident was distracted?

Answer:

a. The probability that a driver will be in an accident = 0.048 or 4.8%

b. The probability that a driver that was in an accident was distracted is = 0.625 or 62.5%

Explanation:

The chance of drivers having an accident due to distraction = 30% or 0.3

Therefore, the chance of not having an accident even with distraction = 70% or 0.7.

The chance of a driver being distracted ° 10% or 0.1

Again, the chance of drivers having an accident when not distracted = 2/100 = 0.02

Therefore the chance of a driver not having an accident and without distraction = 98/100 or 0.98

The chance of a driver not being distracted = 100% - 10% = 90% or 0.9

a. We are required to find the probability that a driver will be in an accident.

Chance of being distracted = 0.1 and had an accident = 0.3

multiplying the two chances = 0.3 × 0.1 = 0.03

Chance of not being distracted = 0.9 and still had an accident = 0.02

multiplying the two chances = 0.9 × 0.2 = 0.018

We will then add the two chances = 0.03 + 0.018 = 0.048 or 4.8%(this is the probability that a driver will be in an accident).

b. We are to calculate the probability that a driver that was in an accident was distracted.

Now, the probability that an accident occured due to distraction = 0.03 and the probability that a driver will be in an accident = 0.048

The chance of an accident taking place as a result of distraction will now be compared with the probability of a driver being in any accident at all for whatever reason.

The probability that a driver that was in an accident was distracted will be obtained by dividing the chance that a driver will be in an accident due to distraction by the probability that a driver will be in an accident:

= 0.03/0.048 = 0.625 or 62.5%

Answer 2

Complete question is;

The police department is keeping track of distracted drivers and accidents. They have found that if a driver is distracted, the driver has a 30% chance of being in an accident. If the driver is not distracted, the driver has a 2% chance of being in an accident. The probability of a driver being distracted is 10%. If needed, create a tree diagram on a separate piece of paper. Then use the diagram to answer the questions.

a. What is the probability a driver will be in an accident? Explain.

b. What is the probability that a driver who was in an accident was distracted? Explain

Answer:

A) 4.8%

B)62.5%

Explanation:

I've attached the tree diagram

A) we are told that the probability of the driver being distracted is 0.1 and the probability of the driver having an accident is 0.3

Thus, probability that the driver is both distracted and may have an accident will be;

P(D&A) = 0.10 x 0.30 = 0.03

Now, since probability that the driver is distracted is 0.1,thus we can say that the probability that he wasn't distracted is; 1 - 0.1 = 0.90

From the question, we are told that if he is not distracted, the probability of having an accident is 0.02.

Thus,the probability of both him not being distracted and having an accident is;

= 0.9 x 0.02 = 0.018

Therefore, the probability of the driver being in accident is;

0.03 + 0.018 = 0.048 or 4.8%

B) We want to find out the probability that a driver who was in an accident was distracted.

In this case, we will ignore the possibilities where the driver did not have an accident.

So, we know that the probability that accident happened while distracted is 0.03.

While probability accident happened while not being distracted is 0.018

Now we are to find probability of accident while distracted compared to the total probability of any accident at all.

Thus, it's ;

0.03/(0.03 + 0.018)

= 0.03/0.048 = 0.625 or 62.5%