3. The probability a regularly scheduled train departs Wilmington, DE Station on

Question

3. The probability a regularly scheduled train departs Wilmington, DE Station on time is . The probability it arrives at Back Bay Station in Boston on time is P(A) = 0.82 . The probability it departs and arrives on time is P(D Intersection A) = 0.78

a. Compute the probability that a train departed Wilmington on time given it arrived in Boston on time.

b. Compute the probability that a train arrives on time in Boston given it departed Wilmington on time.

c. Compute the probability that a train arrives on time in Boston given it did not depart Wilmington on time.

d. Compare P(D) and the answer to part (a). In 1-2 sentences, explain why the comparison makes sense.

Explanation / Answer

a)The probability a regularly scheduled train departs Wilmington, DE Station on time is P(D)=0.83

The probability it arrives at Back Bay Station in Boston on time is P(A) = 0.82

The probability it departs and arrives on time is P(D Intersection A) = 0.78

Therefore, the probability that a train departed Wilmington on time given it arrived in Boston on time =P(D|A)

P(D|A)= P(D Intersection A)/P(A) =(0.78/0.82)=0.9512

b)The probability that a train arrives on time in Boston given it departed Wilmington on time.=P(A|D)

P(A|D)=P(A Intersection D)/P(D)=(0.78/0.83)=0.9398

c)The probability that a train arrives on time in Boston given it did not depart Wilmington on time=P(A|D')

P(A|D')=P(A Intersection D')/P(D')=P(A)*P(D')/P(D')=P(A)=0.82

d)P(D)=0.83 and P(D|A)=0.9512 .So the probability that a train departed Wilmington on time given it arrived in Boston on time i.e.P(D|A) is greater than the Probability of departure because as in the first part it is given that the train arrived in Boston on time So the train must have departed Wilmington on time.

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