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Probablility gambling question

  1. Oct 30, 2009 #1
    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 havent dealt with such a large tree yet,

    i think im 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 dont get the right answer, in fact i get something bigger than 1
     
  2. jcsd
  3. Oct 30, 2009 #2

    lanedance

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    Homework Helper

    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
     
  4. Oct 30, 2009 #3
    but i have 2 possible paths, win1-loss2 win2-loss1
    do i add the results?
     
  5. Oct 30, 2009 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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)
     
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