Lisa is flying from Boston to Denver with a connection in Chicago. The probability her first flight leaves on time is 0.25 . If the flight is on time, the probability that her luggage will make the connecting flight is 0.8 , but if the flight is delayed, the probability that the luggage will make it is only 0.65. Suppose you pick her up at the Denver airport and her luggage is not there. What is the probability that her first flight was delayed?

**Answer**

Let D be the event that Lisa’s first flight was delayed and L be the event that her luggage did not make the connecting flight. We want to find P(D | L).

Using Bayes’ theorem, we have:

P(D | L) = P(L | D) * P(D) / P(L)

where:

P(D) = 0.25 (the probability that the first flight leaves on time)

P(L | D) = 0.65 (the probability that her luggage will make the connecting flight if the flight is delayed)

P(L | D’) = 0.8 (the probability that her luggage will make the connecting flight if the flight is on time)

P(L) = P(L | D) * P(D) + P(L | D’) * P(D’)

So,

P(D | L) = 0.65 * 0.25 / (0.65 * 0.25 + 0.8 * 0.75) = 0.33.

Therefore, the probability that Lisa’s first flight was delayed given that her luggage did not make the connecting flight is 0.33.