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Math Help - Combinatorics

  1. #1
    Moo
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    Combinatorics

    Hello,

    I hate hate hate combinatorics :P
    I roughly have the answers of these ones, but I can't find a correct way to explain everything...

    We have a total of 15 balls in a box : 6 red, 5 green and 4 blue.

    1) We assume that we pick balls one by one and any taken balls will be put back in the box. What is the probability of getting, in this order, a red ball, then a green, then a blue one ?
    The way of doing is blurry to me, but I understood it this way : we take a ball and if it's not red, we put it back. And we continue as long as we get a red ball.
    We keep this ball, and we go on.
    The answer is something like \frac{C_6^1 C_5^1 C_4^1}{C_{15}^3}.


    2) We assume then that we don't put back the balls once they're taken, and we're looking for the same probability.
    I must say I have no idea for this one, I struggle to represent the situation...

    Please help This is no assignment or anything, I'm just helping someone and got stuck to these questions
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    I hate hate hate combinatorics :P
    I feel ya! (an idiom...just in case it is misunderstood)

    I roughly have the answers of these ones, but I can't find a correct way to explain everything...

    We have a total of 15 balls in a box : 6 red, 5 green and 4 blue.

    1) We assume that we pick balls one by one and any taken balls will be put back in the box. What is the probability of getting, in this order, a red ball, then a green, then a blue one ?
    The way of doing is blurry to me, but I understood it this way : we take a ball and if it's not red, we put it back. And we continue as long as we get a red ball.
    We keep this ball, and we go on.
    The answer is something like \frac{C_6^1 C_5^1 C_4^1}{C_{15}^3}.
    I don't think you are interpreting the problem correctly. the problem never said you had to keep picking until you ended up with the "combination". It simply says, if you were to pick 3 balls (with replacement) what is the probability that you will pick a red, then a green, then a blue.

    The probability of picking a red ball is: \displaystyle \frac {C_1^6}{C_1^{15}} = \frac 6{15}

    that is, you can choose 1 of 6 possible choices as there are 6 red balls, out of 15 total choices. so there are 6 ways out of 15 to get a red ball.

    Since the ball is replaced, when choosing the next ball, you still have 15 total choices, of which you must choose 1. So the probabilities for choosing a green and blue ball are: \displaystyle \frac {C_1^5}{C_1^{15}} = \frac 5{15} and \displaystyle \frac {C1^4}{C_1^{15}} = \frac 4{15}, respectively.

    Since each choice is independent, the probability of all three happening is just the product of the probabilities, that is: \displaystyle \frac {C_1^6}{C_1^{15}} \cdot \frac {C_1^5}{C_1^{15}} \cdot \frac {C_1^4}{C_1^{15}} = \frac {C_1^6C_1^5C_1^4}{(C_1^{15})^3}

    2) We assume then that we don't put back the balls once they're taken, and we're looking for the same probability.
    I must say I have no idea for this one, I struggle to represent the situation...

    Please help This is no assignment or anything, I'm just helping someone and got stuck to these questions
    For "without replacement": after picking each ball, the total remaining choices decrease by 1. So after picking the red, there are 14 balls from which to choose the green, then 13 from which to choose the blue. hence this probability is: \displaystyle \frac  {C_1^6C_1^5C_1^4}{C_1^{15}C_1^{14}C_1^{13}}
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Jhevon View Post
    I feel ya! (an idiom...just in case it is misunderstood)

    I don't think you are interpreting the problem correctly. the problem never said you had to keep picking until you ended up with the "combination". It simply says, if you were to pick 3 balls (with replacement) what is the probability that you will pick a red, then a green, then a blue.

    The probability of picking a red ball is: \displaystyle \frac {C_1^6}{C_1^{15}} = \frac 6{15}

    that is, you can choose 1 of 6 possible choices as there are 6 red balls, out of 15 total choices. so there are 6 ways out of 15 to get a red ball.

    Since the ball is replaced, when choosing the next ball, you still have 15 total choices, of which you must choose 1. So the probabilities for choosing a green and blue ball are: \displaystyle \frac {C_1^5}{C_1^{15}} = \frac 5{15} and \displaystyle \frac {C1^4}{C_1^{15}} = \frac 4{15}, respectively.

    Since each choice is independent, the probability of all three happening is just the product of the probabilities, that is: \displaystyle \frac {C_1^6}{C_1^{15}} \cdot \frac {C_1^5}{C_1^{15}} \cdot \frac {C_1^4}{C_1^{15}} = \frac {C_1^6C_1^5C_1^4}{(C_1^{15})^3}

    For "without replacement": after picking each ball, the total remaining choices decrease by 1. So after picking the red, there are 14 balls from which to choose the green, then 13 from which to choose the blue. hence this probability is: \displaystyle \frac  {C_1^6C_1^5C_1^4}{C_1^{15}C_1^{14}C_1^{13}}
    Gosh, all those C's make this look far more complicated than it is! What would Gauss have said - Notation is more important than notions?

    Probability that the first ball is red is 6/15=2/5.

    With replacement the probability that the second is green is 5/15=1/3

    and With replacement the probability that the third is ble is 4/15.

    So the probability of choosing red, green and blue in that order is (2/5)(1/3)(4/15)=8/225

    (Without replacement is only marginally more complicated. With replacement each draw is independednt, without replacement we need to use conditional probabilities)

    CB
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  4. #4
    Moo
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    Well I did the solution with what CB says, but as I misunderstood the what was happening, I hade many difficulties for the second question, using this method.
    The answer provided was in terms of C's, so I guess they want you to do it in a certain way...

    Thanks to both of you I really appreciate it !
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Moo View Post
    Hello,

    I hate hate hate combinatorics :P
    I roughly have the answers of these ones, but I can't find a correct way to explain everything...

    We have a total of 15 balls in a box : 6 red, 5 green and 4 blue.


    2) We assume then that we don't put back the balls once they're taken, and we're looking for the same probability.
    I must say I have no idea for this one, I struggle to represent the situation...

    Please help This is no assignment or anything, I'm just helping someone and got stuck to these questions
    let r,g,b denote drawing ball of colour red, green and blue in that order. and r,g denote drawing red and green in that order ... then:

    P(r,g,b)=P(b|r,g)p(r,g)=(4/13) p(r,g)=(4/13) (5/14) p(r)=(4/13) (5/14) (6/15)

    Ain't conditional probabilities wonderfull things, is there anything they can't do!

    CB
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