Question

You’re a floor manager in a casino. You’ve got two security guards, Smith and Jones, working for you who watch for cheaters. On any given day, about 5% of the people coming in to play cards are cheating. Smith is 88% accurate at detecting cheaters, and Jones is 91% accurate at detecting cheaters. They have been watching a guy in a black baseball hat play cards. Smith says he’s cheating. What’s the probability that Black Baseball Hat is cheating?

Then, after Smith says that Black Baseball Hat is cheating, Jones watches him and says that he’s not. Now what’s the probability that Black Baseball Hat is cheating?  

Answer 1

events :

C : cheater

NC : not cheater

S : smith says he's cheater

NS : smith says he's not cheater

J : jones says he's cheater

NJ : jones says he's not cheater

given : P(C) = 0.05 , P(NC) = 0.95 P(smith correct) = 0.88 , P(smith incorrect) = 1-0.88 P(jones correct) = 0.91 , P(jones incorrect) = 1-0.91

a.

P(S | C) = 0.88 , P(S | NC) = 0.12

P(J | C) = 0.91 , P(J | NC) = 0.09

bayes theorem :

P(S) = P(S | C)*P(C) P(S | NC)*P(NC)

= 0.88*0.05 + 0.12*0.95 = 0.158

P(C | S) = P(S | C)*P(C) / P(S)

= 0.88*0.05 / 0.158

P(C | S) = 0.2785

b.

P(S , NJ) = P(C)*P(Smith correct)*P(jones incorrect) + P(NC)*P(smith incorrect)*P(jones correct)

= 0.05*0.88*(1-0.91) + 0.95*(1-0.88)*0.91

= 0.1077

P(C | S,NJ) = P(C)*P(Smith correct)*P(jones incorrect) / P(S , NJ)

= 0.05*0.88*(1-0.91) / 0.1077

P(C | S,NJ) = 0.0368

P.S. (please upvote if you find the answer satisfactory)