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.

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 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.

kropot72 · Answer

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a. From the tree diagram, the probability of a distracted driver being in an accident is 
$$P(distracted/accident)=0.1\times 0.3=0.030$$
ans the probability of a driver who was not distracted being in an accident is
$$P(undistracted/accident)=0.9\times 0.02=0.018$$
The probability a driver will be in an accident is the sum of the two probabilities calculated above.

kropot72 · Answer

b. The probability that a driver who was in an accident was distracted is found from
$$\frac{P(distracted/accident}{P(accident)}=\frac{0.03}{0.03+0.018}$$

anonymous · Answer

Thanks @kropot72 , since the site was down I was forced to figure it out on my own, but I got it, thanks anyways!

kropot72 · Answer

You're welcome :)