Answer to Question #336623 in Statistics and Probability for curties

Question

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?

Answer 1

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

H_{1 } - he has played roulette,

H_{2 } - he has played blackjack.

We have

$P(H_1)=0.4,P(H_2)=0.6,\\ P(A|H_1)=0.3,P(A|H_2)=0.2.$

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

Using Bayes’ theorem formula we get:

$P(H_2|A)=\\ =\cfrac{P(A|H_2)P(H_2)}{P(A|H_1)P(H_1)+P(A |H_2)P(H_2)}=\\ =\cfrac{0.2\cdot0.6} {0.3\cdot0.4+0.2\cdot0.6} =0.5.$