Question

John flies frequently and likes to upgrade his seat to first class. He has determined that if he checks in for his flight at least 2 hour early, the probability that he will upgrade is 0.8; otherwise, the probability that he will upgrade is 0.3. With his busy schedule, he checks in at least 2 hours before his flight only 40% of the time. At one randomly selected trip that John upgraded his seat to first class, what is the probability that he checked in at least 2 hour before the flight?

Answer 1

Given the conditions, there is a 64% chance that John checked in at least 2 hours early on a randomly selected trip where he upgraded his seat to first class.

Step-by-step explanation:

The student's question is asking for the conditional probability, given that John upgraded his seat to first class, what is the probability that he checked in at least 2 hours before the flight. This can be calculated using Bayes' theorem.

Let's denote A as the event 'John checks in at least 2 hours early', and B as the event 'John upgrades his seat to first class'. We are given that P(A) = 0.4 (he checks in early 40% of the time), P(B|A) = 0.8 (the probability of an upgrade given that he checks in early), and P(B|not A) = 0.3 (the probability of an upgrade if he does not check in early).

The total probability of upgrading, P(B), can be calculated using the law of total probability: P(B) = P(B|A)*P(A) + P(B|not A)*P(not A). Plugging in the given values, we get P(B) = 0.8*0.4 + 0.3*0.6 = 0.50.

Now, using Bayes' theorem we can find P(A|B), the probability that he checked in early given that an upgrade occurred:

P(A|B) = (P(B|A)*P(A)) / P(B) = (0.8*0.4) / 0.50 = 0.64.

So, there is a 64% chance that John checked in at least 2 hours early given that he upgraded his seat to first class.

Answer 2

Answer: Our required probability is 0.64.

Step-by-step explanation:

Since we have given that

Let A be the event that he upgrades his seat to first class.

Let B be the event that he checks in at least 2 hours early.

Let B' be the event that he does not checks in at least 2 hours early.

So,

`P(A|B)=0.8\\\\P(A|B')=0.3\\\\P(B)=40\%=0.4\\\\P(B')=60\%=0.6`

By using Bayes' theorem, we get that

`P(B|A)=(P(B).P(A|B))/(P(B).P(A|B)+P(B').P(A|B'))\\\\P(B|A)=(0.8* 0.4)/(0.8* 0.4+0.3* 0.6)\\\\P(B|A)=(0.32)/(0.32+0.18)\\\\P(B|A)=0.64`

Hence, our required probability is 0.64.