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

Can anyone walk me step by step through this? I’m really having trouble with this unit and would really appreciate it!

### 2 Answers

First draw a probability tree. Start with branches for distracted and not distracted. Then branch off of there for accident or no accident. Be sure to include the percentages for each branch, as a decimal:

…………… …………. / Accident (0.30)

../ Distracted (0.10)

./ …………. …………. No accident (0.70)

/

. …………… …………… / Accident (0.02)

.. Not distracted (0.90)

………………. …………… No accident (0.98)

PART A:

There are two cases.

Driver was distracted (0.10) and had an accident (0.30)

0.10 x 0.30 = 0.03

Driver was not distracted (0.90) and had an accident (0.02)

0.90 x 0.02 = 0.018

Add these together:

0.03 + 0.018

= 0.048

Answer:

4.8%

PART B:

This is a little trickier. Here we can ignore the cases where the driver did not have an accident. We have only two cases, as we figured out in part A.

0.03 –> accident happened while distracted

0.018 –> accident happened while not distracted

We want the first case (accident while distracted) compared to the total probability of any accident (two cases added together).

So basically it is:

0.03

0.03 + 0.018

We already figured out the denominator from part A:

0.03

0.048

That reduces to:

30 .. 5

— = — = 0.625

48 .. 8

Answer:

62.5% a driver that was in an accident was distracted

Hope you understand better.