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.