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Combination problem without replacement

  1. Jul 18, 2008 #1
    1. The problem statement, all variables and given/known data
    a box contains 4 black balls and 6 white balls. After randomly selecting 2 balls with out replacement, what is the probability that both are the same color.


    2. Relevant equations

    C(n,k)=(n!)/(k!(n-k)!)



    3. The attempt at a solution
    I'm just guessing here but,

    P=(C(4,2)C(6,2))/C(10,2)= (6x5x4x3)/(2x10x9)

    The result makes sense I guess. I mean, the probability of picking 2 black balls is (6x5)/(10X9) because the first pick generates a probability of 6/10 and the second pick is out of 9 balls and the possible events is the first 6 times the remaining 5. The same goes for the white balls. And using that logic, it makes sense that the probability is just all the possible events multiplied divided by the twice the overall outcomes( twice because there are 9x10 outcomes for picking 2 of either white or black so twice when the whole picture is taken into account? IDK.) Is my logic correct or did I screw this up?
     
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  3. Jul 18, 2008 #2

    Dick

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    Your value of P is larger than 1. That should tell you something. The numerator should be the number of ways of choosing two white balls PLUS the number of ways of choosing two black balls (and the denominator is the number of ways of choosing any two balls). So don't you want C(6,2)+C(4,2) in the numerator?
     
  4. Jul 18, 2008 #3
    I was thinking about adding them. But I don't know why.
     
  5. Jul 18, 2008 #4

    Dick

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    Because you are trying to count the total number of ways to successfully choose two balls of the same color. C(6,2) for one color, C(4,2) for the other.
     
  6. Jul 18, 2008 #5
    ah, I see. So when dealing with the same color, you multiply right? For example, if I wanted to find the probability of getting 2 of the same color in a row, then it would be, for black (4/10)(3/9) or C(4,2)/C(10,2).
     
  7. Jul 18, 2008 #6

    Dick

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    Sure, and for white probability C(6,2)/C(10,2). Adding that to black, total (C(4,2)+C(6,2))/C(10,2).
     
  8. Jul 18, 2008 #7
    Ok another thing that I'm confused about is multiplying among different sets. For example, If I want to find the probability of getting a white and a black. I would just multiply them, (6/10)(4/9) or (4/10)(6/9) depending on the order. But the problem is that I can also use combinations. (C(4,1)C(6,1))/C(10,2)=(2x6x4)/(9x10) which is not equal to what I have earlier.
     
  9. Jul 18, 2008 #8

    Dick

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    The probability of both the same is 7/15. Not the same then must be 8/15. C(4,1)C(6,1))/C(10,2)=8/15. Good so far. (6/10)(4/9)=4/15. (4/10)(6/9)=4/15. If I add the two order possibilities I still get 8/15. I don't see the problem. The difference is that in the first case you need two black OR two white. In the second case you need one black AND one white. OR=+, AND=*.
     
  10. Jul 18, 2008 #9
    I'm kinda confused by this statement. My logic is that since( say I pick black first) I pick black, the probability would be 6/10, but since there is only 9 balls left, the chance of getting 1 white out of 9 remaining balls is 4/9. So this means that I can interpret this picking black AND picking white. Presumably the order doesn't matter since picking black first and then white is the same as white and then black. Sequentially, it would be 6ways times 4ways all over 10 ways times 9 ways. that why it think its weird that C(4,1)C(6,1))/C(10,2)=/= 4/15
     
  11. Jul 18, 2008 #10

    Dick

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    If you pick black first the odds are 4/15. If you pick white first the odds are 4/15. The two cases are mutually exclusive. You can't do them both at the same time. You have to add them. If you calculate things that way, you are assuming an order. So order does matter.
     
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