Question

Alex 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 0.95 ​, but if the flight is​ delayed, the probability that the luggage will make it is only 0.55 . Suppose you pick her up at the Denver airport and her luggage is not there. What is the probability that ​'s first flight was​ delayed?

​P(delayed)
​(Do not round until the final answer. Then round to three decimal places as​ needed.)

Answer 1

Answer: the probability that Alex's first flight was delayed given that her luggage did not make it to Denver is approximately 0.934 (rounded to three decimal places).

Explanation:

Let D denote the event that the first flight is delayed, and L denote the event that Alex's luggage makes the connecting flight. We want to find P(D|~L), the probability that the first flight was delayed given that her luggage did not make it to Denver.

We can use Bayes' theorem to calculate this probability:

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

We are given that P(D) = 1 - P(first flight leaves on time) = 1 - 0.25 = 0.75, and that P(L|D) = 0.55 and P(L|~D) = 0.95. Therefore, we can calculate P(~L) using the law of total probability:

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

= (1 - P(L|D)) * 0.75 + (1 - P(L|~D)) * 0.25

= 0.3625

Substituting these values into Bayes' theorem, we get:

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

= 0.45 * 0.75 / 0.3625

= 0.934

Therefore, the probability that Alex's first flight was delayed given that her luggage did not make it to Denver is approximately 0.934 (rounded to three decimal places).