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Can someone help with this problem?  I'm totally stuck.

 

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

 Feb 18, 2020
 #1
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+1

 

                                    .02   (accident)

.90 (not distracted) 

                                    .98(no accident)

 

 

                                     .30(accident)

.10 (distracted)

                                     .70 (no accident)

 

 

a.  P(accident) =  P ( not distracted and accident)  + P ( distracted and  accident)  =

             

                                   (.90) (.02)  +  (.10) (.30)  =

 

                                         0.048  = 

 

                                          4.8%

 

 

b. P ( distracted and accident)  =   (.30)(.10)  = 0.03  = 3%

 

 

cool cool cool

 Feb 18, 2020

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