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Math Help - No of selections and probability

  1. #1
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    No of selections and probability

    Six balls are selected from a set of 44 balls, numbered from 1 to 44. The balls are colored. The balles numbered, 1, 5, 9 ... are red, 2,6,10 are orange, the balls no, 3 ,7, 11 are yellow and the balls numbered 4,8,12... are pink

    What is the probability that the selected six balls contain:
    f)Four colours:

    Heres my attempt: there are two cases:
    [1,1,1,3] and [1,1,2,2]

    This equates to
    (4C3 x (11C1)^3 x (11C3) + 4C2 x (11C1)^2 x (11C2)^2)/44C11

    However the book has a different solution:
    (4C3 x (11C1)^3 x (11C3) + 4C2 x 2C1 X (11C1)^2 x (11C2)^2)/44C11

    Can someone explain to me?
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  2. #2
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    Quote Originally Posted by Lukybear View Post
    Six balls are selected from a set of 44 balls, numbered from 1 to 44. The balls are colored. The balles numbered, 1, 5, 9 ... are red, 2,6,10 are orange, the balls no, 3 ,7, 11 are yellow and the balls numbered 4,8,12... are pink

    What is the probability that the selected six balls contain:
    f)Four colours:

    Heres my attempt: there are two cases:
    [1,1,1,3] and [1,1,2,2]

    This equates to
    (4C3 x (11C1)^3 x (11C3) + 4C2 x (11C1)^2 x (11C2)^2)/44C11

    However the book has a different solution:
    (4C3 x (11C1)^3 x (11C3) + 4C2 x 2C1 X (11C1)^2 x (11C2)^2)/44C11

    Can someone explain to me?
    shouldn't the denominator be 44C6?

    the 2C1 is there because {1 red, 1 orange, 2 yellow, 2 pink} is different from {2 red, 2 orange, 1 yellow, 1 pink}.
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  3. #3
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    Yea it should be 44C6. But isnt the 2C1 counted for by 4C2?
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  4. #4
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    Quote Originally Posted by Lukybear View Post
    Yea it should be 44C6. But isnt the 2C1 counted for by 4C2?
    This is under case {1,1,2,2}

    4C2 means you're selecting two objects from four. Think about what exactly it is you are selecting. Say you select red and orange. Do you assign these to {1,1} or to {2,2}? It could be either. Therefore you multiply by 2 which equals 2C1.
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  5. #5
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    Yes that makes more sense.Thanks
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  6. #6
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    Hmm. After it made sense, it had seemed that the answer was wrong after all. There is no 4C2 . Can you help me make sense of it now.
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  7. #7
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    Quote Originally Posted by Lukybear View Post
    Hmm. After it made sense, it had seemed that the answer was wrong after all. There is no 4C2 . Can you help me make sense of it now.
    I am rather puzzled. I would like to write a brute force computer program (since 44C6 is pretty small) and compare results, then get to the bottom of it. I may or may not have time to do that right now.
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  8. #8
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    Quote Originally Posted by undefined View Post
    I am rather puzzled. I would like to write a brute force computer program (since 44C6 is pretty small) and compare results, then get to the bottom of it. I may or may not have time to do that right now.
    Oh all right, as often happens I have made a silly error. Sorry about that.

    Your original answer is right, and the book's answer is wrong.

    Once you select the two colours to correspond to (for example) {1,1} then the colours for {2,2} are fixed, and by multiplying by 2 you double count.

    Here's the Java program I wrote

    Code:
    import java.util.ArrayList;
    
    public class ColouredBallsProb {
        static int c;
        
        public static void main(String[] args) {
            long t=time();
            c=0;
            recurse(0, new ArrayList<Integer>());
            System.out.println(c);
            System.out.println("elapsed: "+(time()-t)/1000.0+" s");
        }
        
        static void recurse(int ind, ArrayList<Integer> balls) {
            if(balls.size()==6) {
                if (oneOfEach(balls)) c++;
                return;
            }
            if(ind>43) return;
            // select or don't select current ball
            recurse(ind+1,balls);
            ArrayList<Integer> nextBalls=new ArrayList<Integer>(balls);
            nextBalls.add(ind);
            recurse(ind+1,nextBalls);
        }
        
        static boolean oneOfEach(ArrayList<Integer> balls) {
            // assume 6 balls
            int i;
            boolean[] colours=new boolean[4];
            for (i=0;i<6;i++) {
                colours[balls.get(i)%4]=true;
            }
            for (i=0;i<4;i++)
                if(!colours[i])
                    return false;
            return true;
        }
        
        static long time() {
            return System.currentTimeMillis();
        }
    }
    Output

    Code:
    3074610
    elapsed: 1.101 s
    As you can see it matches the first answer you gave (just the numerator). Sorry for the mix up.
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  9. #9
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    Wow. That is brilliant. Absolutely fantastic.

    I get your fixing explanation. But what about multiplying by 2? Wheres that?
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  10. #10
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    Quote Originally Posted by Lukybear View Post
    Wow. That is brilliant. Absolutely fantastic.

    I get your fixing explanation. But what about multiplying by 2? Wheres that?
    I don't follow.. The whole issue is whether the 2C1 in bold in post #1 should appear in the answer right? It shouldn't.

    When you wrote 4C2 in post #6 I assumed it was a typo and was supposed to be 2C1.

    The answer is

    (4C3 x (11C1)^3 x (11C3) + 4C2 x (11C1)^2 x (11C2)^2)/44C6

    Hope it's clear now?
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  11. #11
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    Yea thanks.
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