Bill commutes to work in the business district of Boston. He takes the train 3/5 of the time and drives 2/5 of the time (when he visits clients). If he takes the train, then he gets home by 6:30 p. m. 83% of the time; if he drives, then he gets home by 6:30 p. m. 62% of the time. If Bill gets home by 6:30 p. m., what is the probability that he drove to work? (Round your answer to three decimal places.)

The probability that Bill drove to work given that he reached home by 6:30 PM is 0.3324.

Step-by-step explanation:

Denote the events as follows:

T = Bill takes the train to work

D = Bill drives to work

X = Bill gets home from work by 6:30 PM.

The information provided is:

P (T) = 0.60

P (D) = 0.40

P (X|T) = 0.83

P (X|D) = 0.62

The Bayes's theorem states that the conditional probability of an event Ei given that another event has already occurred is:

Use the Bayes' theorem to determine the value of P (D|X) as follows:

Thus, the probability that Bill drove to work given that he reached home by 6:30 PM is 0.3324.

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

step-by-step explanation: