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Math Help - Choosing 5 balls from 10 black, 20 white, 40 red.

  1. #1
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    Choosing 5 balls from 10 black, 20 white, 40 red.

    10 black, 20 white, 40 red

    The question is what is probability when we reach 5 of them, and we have 2 white, 2 red, 1 black.......
    And some guy solved this question on youtube, and his answer was 12.2%
    He solve like : (20C2 x 40C2 x 10C1)/ (70C5)


    I know he was right. BUt I don't understand it

    I solve by another way like :
    Doing it your way, you would calculate the probability for each of the 30 possible ways to draw 2W, 2R and one B. So, to draw WWRRB in that order, you get
    (20/70)(19/69)(40/68)(39/67)(10/66)

    Each iteration of 2W, 2R, and 1B has an identical probability. So, multiply by 30, and you get 12.2%.

    Can somebody explain me about first way?
    I really can't make sense!!
    Last edited by mr fantastic; June 10th 2011 at 07:36 PM. Reason: Re-titled.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by daivinhtran View Post
    10 black, 20 white, 40 red

    The question is what is probability when we reach 5 of them, and we have 2 white, 2 red, 1 black.......
    And some guy solved this question on youtube, and his answer was 12.2%
    He solve like : (20C2 x 40C2 x 10C1)/ (70C5)


    I know he was right. BUt I don't understand it

    I solve by another way like :
    Doing it your way, you would calculate the probability for each of the 30 possible ways to draw 2W, 2R and one B. So, to draw WWRRB in that order, you get
    (20/70)(19/69)(40/68)(39/67)(10/66)

    Each iteration of 2W, 2R, and 1B has an identical probability. So, multiply by 30, and you get 12.2%.

    Can somebody explain me about first way?
    I really can't make sense!!
    Do you know what nCk means?

    You have to choose 5 balls from 70(10+20+40), you have 70C5 combinations , now there some conditions:

    i. 2 W balls from 20 possible. (20C2)

    ii. 2 R balls from 40 possible. (40C2)

    iii. 1 black ball from 10 possible. (10C1)

    Can you proceed?
    Last edited by Also sprach Zarathustra; June 10th 2011 at 08:20 PM.
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  3. #3
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Do you know what nCk means?

    You have to choose 5 balls from 70(10+20+40), you have 70C5 combinations , now there some conditions:

    i. 2 W balls from 20 possible. (20C2)

    ii. 2 R balls from 40 possible. (40C2)

    iii. 1 black ball from 10 possible. (10C1)

    Can you proceed?
    i KNOW WHAT 70c5 MEAN..... my ways, and his way 's so different..... It's not related with his way.... BUt it comes out same answer
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  4. #4
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    When you have an unordered, without replacement, choice, you know the possible orderings. If, say, you were dealing with a poker hand, you would have

    52\times 51\times 50\times 49\times 48

    possible orderings, but you have to divide out the redundant orderings. The five card hand can be arranged in

    5\times 4\times 3\times 2\times 1

    ways, so the total number of unordered hands is

    \frac{52\times 51\times 50\times 49\times 48}{5\times 4\times 3\times 2\times 1} = \frac{52!}{5!47!}

    Now consider what you wrote above:

    \frac{20\times 19\times 40\times 39\times 10}{70\times 69\times 68\times 67\times 66}\times 30

    See something similar? Now write out the statements like \binom{70}{5} and the other ones. With a little algebra, I'm sure you will find that one expression does, in fact, lead to the other.

    Spoiler:

    \frac{\binom{20}{2}\times \binom{40}{2}\times \binom{10}{1}}{\binom{70}{5}}

    =\frac{\frac{20\times 19\times 40\times 39\times 10}{4}}{\frac{70\times 69\times 68\times 67\times 66}{5\times 4\times 3\times 2\times 1}}

    =\frac{20\times 19\times 40\times 39\times 10\times 120}{70\times 69\times 68\times 67\times 66\times 4}

    =\frac{20\times 19\times 40\times 39\times 10\times 30}{70\times 69\times 68\times 67\times 66}


    You simply thought about the problem in another way, but that way misses the point of using the binomial coefficient: it simplifies notation and intuition. You may not be comfortable with thinking about it in terms of the choose operation, but you should work at it. Your way develops from the basic counting rules, but so does the choose operation. The advantage of the choose operation is that we can simplify the expression to the first one above, which carries a much simpler intuition: e.g., from choosing 2 whites from a bin of 20 whites, we have 20 choose 2 many. Which is easier to comprehend, that or your accounting for the various orders, that they're all equally likely, and that there are 30 of them? The say the same thing, ultimately, but the choose operation is simpler.
    Last edited by bryangoodrich; June 11th 2011 at 11:53 AM. Reason: solution
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  5. #5
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    Quote Originally Posted by daivinhtran View Post
    10 black, 20 white, 40 red

    The question is what is probability when we reach 5 of them, and we have 2 white, 2 red, 1 black.......
    And some guy solved this question on youtube, and his answer was 12.2%
    He solve like : (20C2 x 40C2 x 10C1)/ (70C5)


    I know he was right. BUt I don't understand it

    I solve by another way like :
    Doing it your way, you would calculate the probability for each of the 30 possible ways to draw 2W, 2R and one B. So, to draw WWRRB in that order, you get
    (20/70)(19/69)(40/68)(39/67)(10/66)

    Each iteration of 2W, 2R, and 1B has an identical probability. So, multiply by 30, and you get 12.2%.

    Can somebody explain me about first way?
    I really can't make sense!!
    The way he did it is by counting the number of ways to get 2 reds with 2 white and 1 black.
    There are 40 reds, so the number of ways to pick a pair of reds is 40C2.
    The number of ways to pick 2 whites from the 20 is 20C2.
    There are of course 10C1 = 10 ways to pick the black.

    In obtaining the group of 5 balls with 2 red, 2 white and 1 black,
    realise that ANY pair of whites can be matched with ANY pair of reds.
    Hence the number of possible pairs of reds is multiplied by the number of possible pairs of whites.
    Then the resulting groups of 4 can all be matched with any of the 10 blacks.
    So we multiply by 10 to find the number of different ways to obtain a group of 2 reds, 2 whites and 1 black.

    Then the total number of such groups of 5 possible are 10C2 X 40C2 X 10C1.

    (If you choose small numbers, it's easier to get used to.
    Suppose there are 3 reds, 4 whites and 2 blacks.
    There are 3 ways to pick a pair of reds, 6 ways to pick a pair of whites and 2 ways to pick a black.
    Therefore, there are 3X6X2 such groups or sets of 5 satisfying the condition.
    That gives the probability numerator.
    The denominator is the number of groups of 5 from the total without any restrictions placed)

    Then the denominator of the probability fraction is the number of ways to pick a group of 5 from 70.

    You could think of this method as "picking all 5 at the same time".

    Your method is more akin to picking them "sequentially" one after the other,
    so you've taken into account the orders.
    You must get the same result.
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