Probablility gambling question

[Dell]

a man walks into a casino and sees 2 slot machines, he randomly chooses one.
the chance of winning on machine 1 is 0.4
the chance of winning on machine 2 is 0.3
if the man wins he plays a second game on the same machine, if he loses he changes to the other machine

what is the probability of him winning one game and losing one game

what i have been doing up till this question was building a tree diagram with the possibilities

A-win in first round
B-win in second round

since he randomly chooses a machine i have a 50/50 chance of playing on either machine

the problem is that i haven't dealt with such a large tree yet,

i think I am looking for P(A/[tex]\bar{B}[/tex])+P([tex]\bar{A}[/tex]/B)+P(B/[tex]\bar{A}[/tex]) +P([tex]\bar{B}[/tex]/A) and i need to do this for each of the 2 machines

but i don't get the right answer, in fact i get something bigger than 1

 

[lanedance]

at each branch node, the sum of the probabiltys on each branch should be 1.

Multiply the probabilty along a path to get the result, which has to be <1 as all the multpliers are <1.

the first choice will be which machine to start at, as its random, it will be P=0.5 for each machine

 

[Dell]

but i have 2 possible paths, win1-loss2 win2-loss1
do i add the results?

 

[HallsofIvy]

There are four paths to consider, not 2:
(1) Win on machine 1, then lose on machine 1
(2) Lose on machine 1, then win on machine 2
(3) Win on machine 2, then lose on machine 2
(4) Lose on machine 2, then win on machine 1

The probability for (1) is (.4)(.6) and the probability of (2) is (.6)(.3). Since the probability of starting on machine 1 is .5, add those two numbers and multiply by .5.

Do the same for (3) and (4)