Help with a conditional probability problem There are 6 balls in a bag and they are numbered 1 to 6.
We draw two balls without replacement.
Is the probability of drawing a "6" followed by drawing an "even" ball the same as the probability of drawing an "even" ball followed by drawing a "6".
According to Bayes Theorem these two possibilities should be the same:
Pr(A and B) = Pr(A) x Pr(B∣A)
Pr(A and B) = Pr(B) x Pr(A∣B)
However, when I try to work this out I am getting two different probabilities, 2/30 and 3/30 for the two different scenarios listed above. The first scenario is fairly straight-forward to determine,
Pr(6) x Pr(even∣6 has already been drawn)
1/6 x 2/5 = 2/30
however, I think I am doing something wrong with the second scenario,
Pr(even) x Pr(6∣even has already been drawn)
3/6 x ?????
Any help would be greatly appreciated as this is really bugging me.
Thank you in advance....

There is no need to compute anything. All orders of drawing the balls are equally likely. 
