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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?

Respuesta :

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,

[tex]P(A|B)=0.8\\\\P(A|B')=0.3\\\\P(B)=40\%=0.4\\\\P(B')=60\%=0.6[/tex]

By using Bayes' theorem, we get that

[tex]P(B|A)=\dfrac{P(B).P(A|B)}{P(B).P(A|B)+P(B').P(A|B')}\\\\P(B|A)=\dfrac{0.8\times 0.4}{0.8\times 0.4+0.3\times 0.6}\\\\P(B|A)=\dfrac{0.32}{0.32+0.18}\\\\P(B|A)=0.64[/tex]

Hence, our required probability is 0.64.

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