Question #336623

A gambler plays only one game for at most one hour each evening. The chance of showing a profit at the end of the hour if he plays roulette and blackjack is 30% and 20% respectively. He chooses blackjack on 60% of the evenings and roulette on the remaining evenings. Suppose he shows a profit after playing one a particular evening. What is the probability that that he played blackjack?


1
Expert's answer
2022-05-03T16:16:04-0400

Let A be an event that the gambler shows a profit after playing,

H1  - he has played roulette,

H2  - he has played blackjack.

We have

P(H1)=0.4,P(H2)=0.6,P(AH1)=0.3,P(AH2)=0.2.P(H_1)=0.4,P(H_2)=0.6,\\ P(A|H_1)=0.3,P(A|H_2)=0.2.


Then we are to find the value of P(H2A)P(H_2|A) .

Using Bayes’ theorem formula we get:

P(H2A)==P(AH2)P(H2)P(AH1)P(H1)+P(AH2)P(H2)==0.20.60.30.4+0.20.6=0.5.P(H_2|A)=\\ =\cfrac{P(A|H_2)P(H_2)}{P(A|H_1)P(H_1)+P(A |H_2)P(H_2)}=\\ =\cfrac{0.2\cdot0.6} {0.3\cdot0.4+0.2\cdot0.6} =0.5.





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