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Math Help - Balls in a bag

  1. #1
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    Balls in a bag

    There are 10 red balls, 10 green balls and 6 white balls.
    Two balls picked at random - what is the probability that they are of different colors?

    so probability of the first ball is 1/26 (if i'm right?), i don't get how to find t he probablity of the second one?
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  2. #2
    Super Member Quacky's Avatar
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    The probability of each ball being picked is \frac{1}{26}, which is a start.

    The way I would solve this would be to find the probability of two red balls, the probability of two green balls, and the probability of two white balls. If you add these, you will find the probability of receiving two balls that are the same colour.

    I'm going to assume (as you haven't specified) that the balls are not replaced, and that this is therefore conditional probability.

    P(2 red) = \frac{10}{26}\times\frac{9}{25} = \frac{9}{65}. This is because there are 10 red balls out of the 26 total during the first selection, and then there are 9 red balls left out of the 25 remaining balls during the second selection.
    P(2 green) is the same as p(2 red) because there are the same amount of red balls as green balls, so p(2 green) =\frac{9}{65}
    P(2 white) = \frac{6}{26}\times\frac{5}{25}=\frac{3}{65}

    Do you understand so far?

    So what is the total probability of receiving two balls that are the same colour?

    And, therefore, how can you work out from that the probability of receiving two balls which are not the same colour? Show your working if you get stuck.
    Last edited by Quacky; March 3rd 2011 at 11:42 AM.
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  3. #3
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    Quote Originally Posted by kandyfloss View Post
    There are 10 red balls, 10 green balls and 6 white balls.
    Two balls picked at random - what is the probability that they are of different colors?

    so probability of the first ball is 1/26 (if i'm right?), i don't get how to find t he probablity of the second one?
    Draw a tree diagram.
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  4. #4
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    So i did (1-P(sum of all of the probabilities)) and got 44/65. That seems right.
    Thank you.
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  5. #5
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    Quote Originally Posted by kandyfloss View Post
    There are 10 red balls, 10 green balls and 6 white balls. Two balls picked at random - what is the probability that they are of different colors?
    Probability both are white is P(W_1\cap W_2)=\frac{6}{26}\cdot\frac{5}{25}.

    So find P(R_1\cap R_2)+P(G_1\cap G_2)+P(W_1\cap W_2)
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  6. #6
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    Quote Originally Posted by kandyfloss View Post
    So i did (1-P(sum of all of the probabilities)) and got 44/65. That seems right.
    Thank you.
    You could have tried...

    What is the probability of getting

    (1) a red with a green

    or

    (2) a red with a white

    or

    (3) a green with a white


    There are \binom{26}{2} ways to choose 2 balls. This is the denominator for your probability fraction.

    There are 10(10)=100 ways to get a green with a red

    There are 10(6)=60 ways to get a red with a white

    There are 10(6)=60 ways to get a green with a white

    The sum of the above three products is the numerator of your probability fraction.
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  7. #7
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    I went through, both of your methods. Since I already did it using the first one, i'll stick to it.
    But thanks you very much it really helped me understand better.
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  8. #8
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    [1]Prob white : 6/26 = 3/13 ; [2]prob not white : 20/25 = 4/5 ; 3/13 * 4/5 = 12/65
    [1]Prob not white: 20/26 = 10/13 ; [2]prob not same: 16/25 ; 10/13 * 16/25 = 32/65

    12/65 + 32/65 = 44/65
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