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Math Help - Another ball problem

  1. #1
    Ond
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    Another ball problem

    Bowl A contains 4 red balls, 3 blue balls and 2 green balls. Bowl B contains 2 red balls, 3 blue balls and 4 green balls. One ball is drawn from bowl A and then put into bowl B. After that one ball is drawn from bowl B. What is the probability that the ball that is drawn from bowl B is red?

    Thanks in advance...
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  2. #2
    Ond
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    Quote Originally Posted by Ond View Post
    Bowl A contains 4 red balls, 3 blue balls and 2 green balls. Bowl B contains 2 red balls, 3 blue balls and 4 green balls. One ball is drawn from bowl A and then put into bowl B. After that one ball is drawn from bowl B. What is the probability that the ball that is drawn from bowl B is red?

    Thanks in advance...
    Well, think I've got it, but I'm not quite sure. Is it: 4/9 * 3/10 + 5/9 * 2/10
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  3. #3
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    Quote Originally Posted by Ond View Post
    Bowl A contains 4 red balls, 3 blue balls and 2 green balls. Bowl B contains 2 red balls, 3 blue balls and 4 green balls. One ball is drawn from bowl A and then put into bowl B. After that one ball is drawn from bowl B. What is the probability that the ball that is drawn from bowl B is red?
    \mathcal{P}\left(R_2\right)=\mathcal{P}\left(R_1R_  2\right)+\mathcal{P}\left(B_1R_2\right)+\mathcal{P  }\left(G_1R_2\right)
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  4. #4
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    Well, I'll throw in my 2 cents' worth:

    Prob that 1st ball is red = 4/9

    So we have 2 + 4/9 reds = 22/9 reds

    Prob that next draw is red: (22/9) / (22/9 + 3 + 4) = 22/85

    Anything wrong with this "1st cup of coffee" approach?
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