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Math Help - Probablity Question - have I done this right?

  1. #1
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    Probablity Question - have I done this right?

    Hi there! I'm wondering if I've done the question below right ...

    The teacher shows the students two bags. Bag A contains 3 gold coins and 2 silver coins. Bag B contains 4 gold coins and 5 silver coins. A coin is picked at random from Bag A and placed in Bag B. A coin is then picked at random from Bag B and placed in Bag A. These two operations are performed again in the same order. What is the probability that Bag A now contains five gold coins?

    This is how I did it:
    - In order to end up with 5 coins in A, each of the two swaps has to consist of putting gold coin into A from B and putting a silver coin into B from A.
    - Probability of swaps occurring as described = probability of 1st swap x probability of 2nd swap = (4/9 x 2/5)(3/9 x 1/5) = 8/675

    Does my logic follow? I'm curious to know as I've nowhere to check the answer!
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  2. #2
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    Re: Probablity Question - have I done this right?

    Quote Originally Posted by HelenMc9 View Post
    The teacher shows the students two bags. Bag A contains 3 gold coins and 2 silver coins. Bag B contains 4 gold coins and 5 silver coins. A coin is picked at random from Bag A and placed in Bag B. A coin is then picked at random from Bag B and placed in Bag A. These two operations are performed again in the same order. What is the probability that Bag A now contains five gold coins?

    This is how I did it: In order to end up with 5 coins in A, each of the two swaps has to consist of putting gold coin into A from B and putting a silver coin into B from A.
    - Probability of swaps occurring as described = probability of 1st swap x probability of 2nd swap = (4/9 x 2/5)(3/9 x 1/5) = 8/675
    I have no idea where you are getting those numbers. Where do you get (4/9)?

    In order to end with all gold coins in A we must gave the string S_1G_2S_3G_4.

    That is, first ball is silver from A, second is gold from B, third is silver from A, and fourth is gold from B.

    Find \mathcal{P}(S_1G_2S_3G_4)=\mathcal{P}(S_1)\mathcal  {P}(G_2|S_1)\mathcal{P}(S_3|S_1G_2)\mathcal{P}(G_4  |S_1G_2S_3)
    Last edited by Plato; May 13th 2013 at 04:48 PM.
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