Question

Leah is flying from Boston to Denver with a connection in Chicago. The probability her first flight is on time is 0.15. If the flight is on time, the probability that her luggage will make the connecting flight in Chicago is 0.95, but if the first flight is delayed, the probability that the luggage will make it is only 0.65.

Question: Suppose you pick up Leah in Denver and her luggage is not there.
What is the probability that her first flight was delayed?

Answer 1

Answer: 0.975

Step-by-step explanation:

Let event that First flight on time be denoted as OT and event that a flight is delayed be D;

Therefore: P(OT) = 0.15 and P(D) = 1 - P(OT) = 1 - 0.15 = 0.85

Let event that Luggage will make connecting flight be L and that it won't make it be L',

Therefore: It is given that if the flight is on time, the probability that luggage will make it is 0.95, This is a conditional probability and can be written as;

P(L/OT) = 0.95 and also P(L/D) = 0.65

From this P(L'/OT) = 1 - P(L/OT) = 1 - 0.95 = 0.05

and P(L'/D) = 1 - P(L/D) = 1 - 0.65 = 0.35

The question is asking for the probability that the first flight was delayed (D) provided that her Luggage was not there (L').

This is a conditional probability question;

Set up the formula : P(D/L') = P( D and L')/ P(L')

P(D and L') = P(D) x P(L'/D) = 0.85 x 0.35 = 0.2975

P(L') = P(OT) x P(L'/OT) + P(D) x P(L'/D)

      = 0.15(0.05) + 0.85(0.35)

      = 0.305

Therefore P(L/OT) = 0.297/0.305

                               = 0.975