According to a survey, 54% of all workers read email while on the phone. Suppose

Q: According to a survey, 54% of all workers read email while on the phone. Suppose

4) P(a worker reads email while talking on the phone and writes "to do" lists in meetings) =0.54*0.20=0.108 5)P(a worker does not write personal "to do" lists given that he reads email while talking on the phone) =(1-0.20) =0.80 6)) P(a worker does not write "to do" lists and does read email while talking on the phone)=(1-0.20)*0.54=0.432

ID: 3052535 • Letter: A

Question

According to a survey, 54% of all workers read email while on the phone. Suppose 20% of those who read email while on the phone write 'to do" lists during meetings. A worker is randomly selected. What is the probability that

4) a worker reads email while talking on the phone and writes "to do" lists in meetings?

5) a worker does not write personal "to do" lists given that he reads email while talking on the phone?

6) a worker does not write "to do" lists and does read email while talking on the phone?

Explanation / Answer

P(a worker reads email while talking on the phone and writes "to do" lists in meetings) =0.54*0.20=0.108

5)P(a worker does not write personal "to do" lists given that he reads email while talking on the phone)

=(1-0.20) =0.80

6)) P(a worker does not write "to do" lists and does read email while talking on the phone)=(1-0.20)*0.54=0.432

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According to a survey, 54% of all workers read email while on the phone. Suppose

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