Bayes Theorem Help

Question

Tommy is a football player who suffers an injury during the game. The team doctor looks at the injury and tells Tommy that he needs to run some tests and there is only a 20% chance he will recommend he go back into the game. While waiting for the doctor to run the tests, an assistant coach asks Tommy what the likelihood of him coming back into the game are, to which Tommy answers 85%. What is the probability that Tommy goes to back into the game given that the doctor let him?

 

P(back in game | doc let him)

= P(doc let him | back in game) * P(back in game) / P(doc let him)

= P(doc let him | back in game) * .85 / .2

 

It's that P(B | A) term that is confusing me. Can anybody help?

Answer 1

I would construct the solution as:

P(Tom Plays | Doctor Approves) equals:

0.85-----×---------0.2

P(Tom Plays) × P(Doctor Approves | Tom Plays)

****************************************************************************************

0.85-----×---------0.2-----------------------------------+-----0.15------------×-----------0.8

P(Tom Plays) × P(Doctor Approves | Tom Plays) + P(Tom Benched)×P(Doctor Approves | Tom Benched)

Then write (0.85×0.2) ÷ [(0.85×0.2)+(0.15×0.8)].

Simplify this last as 0.17/(0.17+0.12) or 17/29

or 0.5862068966.