Question

station 7
rachel is flying from boston to denver with a connection in
chicago. the probability that her first flight leaves on time is.1.
if the flight is on time, the probability that her luggage will make
the connecting flight is.95, but if the flight is delayed, the
probability that the luggage will make it is only.65. what is the
probability that if her luggage arrives with her, her flight is on
time?

Explanation:

Step1: Calculate the probability that the flight is on - time and luggage arrives

Let \(A\) be the event that the flight is on - time and \(B\) be the event that the luggage arrives.
The probability that the flight is on - time \(P(A)=0.1\), and the probability that the luggage arrives given the flight is on - time \(P(B|A) = 0.95\).
By the multiplication rule \(P(A\cap B)=P(B|A)P(A)=0.1\times0.95 = 0.095\)

Step2: Calculate the probability that the flight is delayed and luggage arrives

The probability that the flight is delayed \(P(\overline{A})=1 - P(A)=1 - 0.1=0.9\)
The probability that the luggage arrives given the flight is delayed \(P(B|\overline{A})=0.65\)
By the multiplication rule \(P(\overline{A}\cap B)=P(B|\overline{A})P(\overline{A})=0.9\times0.65 = 0.585\)

Step3: Calculate the total probability that the luggage arrives

By the law of total probability \(P(B)=P(A\cap B)+P(\overline{A}\cap B)=0.095 + 0.585=0.68\)

Step4: Use Bayes' theorem

By Bayes' theorem \(P(A|B)=\frac{P(A\cap B)}{P(B)}\)
Substitute \(P(A\cap B) = 0.095\) and \(P(B)=0.68\)
\(P(A|B)=\frac{0.095}{0.68}\approx0.14\)

Answer:

The probability that if her luggage arrives with her, her flight is on time is approximately \(0.14\)