Fun with Bayes theorem! You wish to test out two slot machines (Machine 1 and 2). One is "good", i.e. offers better chances of winning, and the other is "bad". You do not know which is which as both are identical, the probability of playing either the "good" or "bad" is 50%. 
The probability of winning on the “good” machine is 1/2 and the probability of winning on the “bad” machine is 1/3. What is the probability that Machine 2 is “Good” given you won playing on Machine 1?
ANSWER CHOICES


*

*0.3

*0.4

*0.5

*0.6

*0.7


MY SOLUTION SO FAR...

1# Since each machine is equally likely to be the “good” machine we can express this as...
P(M1 is Good)=P(M2 is Bad)=1/2
P(M1 is Good)=P(M2 is Bad)=1/2
P(M1 is Bad)=P(M2 is Good)=1/2
P(M1 is Bad)=P(M2 is Good)=1/2

2# We have also been told the probability of winning for each type of machine
P(Win on M1 | M1 is Good)=1/2
P(Win on M1 | M1 is Bad)=1/3
P(Win on M1 | M1 is Good)=1/2
P(Win on M1 | M1 is Bad)=1/3

3# We can use these probabilities to calculate the probability of losing for each type of machine as well:
P(Loose on M1 | M1 is Good)=1/2
P(Loose on M1 | M1 is Bad)=2/3
P(Loose on M1 | M1 is Good)=1/2
P(Loose on M1 | M1 is Bad)=2/3

4# The probability of M1 being good given that you won on M1 is:
P(M1 is Good | Win on M1)= P(M1 is Good) * [P(Win on M1 | M1 is Good)/P(Win on M1)]= (1/2*1/2) / (1/2*1/2 + 1/3*1/2)=0.6

5# The probability of M1 being bad given that you won on M1 is:
P(M1 is bad | Win on M1)=P(Mi is bad) * [P(Win on M1|M1 is bad)/P(Win on M1)] = (1/2*1/3) / (1/2*1/2 + 1/3*1/2) = 0.4

I don't know how to jump from machine 1's probability to machine 2...
 A: Let's rephrase the question:

What is the probability that machine 2 is good given that you won playing on machine 1?

To the equivalent question:

What is the probability that machine 1 is bad given that you won playing on machine 1?


Let $A$ denote an event of machine 1 being bad.
Let $B$ denote an event of winning on machine 1.
$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac12\times\frac13}{\frac12\times\frac13+\frac12\times\frac12}=\frac25$
A: There's a lot of typing going on, so let's just introduce some notation:


*

*$M_1$... "Machine $1$ is good".

*$M_2$... "Machine $2$ is good".

*$W_1$... "Win on machine $1$".

*$W_2$... "Win on machine $2$".


Now we can describe everything shortly. For example, losing on machine $2$ is simply $\neg W_2$, since losing is the negation of winning.
Now, you need to calculate the probablity that machine $2$ is good given that you won on machine $1$.
So, you need to calculate
$$P(M_2|W_1)$$
Using Bayes off the bat, you get that $$P(M_2|W_1) = P(W_1|M_2) \cdot \frac{P(M_2)}{P(W_1)}$$
So, what do we have out of the three numbers you need?
Well, $P(W_1|M_2)$ is the probability that you will win on machine $1$ if machine $2$ is good. This is simply the probability of winning on a bad machine, which is $\frac13$.
$P(M_2)$ is even simpler to calculate, it's simply $\frac12$.
The last you need to calculate is $P(W_1)$, i.e. the probability that you will win on machine $1$. To do that, let me first just give you a hint:

Remember the law of total probability.

