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Thread: discs in a bag

  1. #1
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    discs in a bag

    a bag contains 20 discs marked with numbers from 1 to 20 inclusive. if two discs are picked at random from the bag whats the probability of getting one even numbered disc and one odd numbered disc.

    there are 10 even and odd numbers so i was thinking the answer would be 10/20 *10/20 =1/4 the answer at the back of the book is 10/19
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  2. #2
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    Re: discs in a bag

    Hey markosheehan.

    You should think about the order of picking things.

    If you pick one even one then the next even one has a lower chance.

    Try listing the different combinations of frequencies for picking even and odd in different orders and add them up.
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  3. #3
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    Re: discs in a bag

    sorry yes so the answer would be 10/20*10/19=5/19 this is not the right answer either
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  4. #4
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    Re: discs in a bag

    The first disc chosen is even or odd ... probability of choosing one or the other is 1. Choosing the first leaves 19 in the bag.

    You're looking for the second disc to be different than the first ... odd if the first was even or even if the first was odd. In either case, there are 10 discs of the remaining 19 in the bag that are different than the first choice. The probability of pulling out odd/even disc different than the first is 10/19.

    the probability that the fisrt two discs drawn are odd/even or even/odd is 1 times 10/19
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  5. #5
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    Re: discs in a bag

    There are initially 20 disks in the bag, 10 of them even. The probability the first disk drawn is even is 10/20= 1/2. There are then 19 disks in the bag, 10 of them odd. The probability the second disk drawn is odd is 10/19. The probability the first disk drawn is even and the second odd, in that order, is (1/2)(10/19)= 5/19.

    Similarly, since there were also 10 odd disks in the bag initially, the probability the first disk drawn is odd is 1/2. Then there are 19 disks left in the bag, 10 of them even. The probability the second disk drawn is even is 10/19. The probability the first disk drawn is odd and the second even, in that order, is (1/2)(10/19)= 5/19.

    The probability one disk is even and the other odd, in either order, is 5/19+ 5/19= 10/19.

    (It is always the case, in problems like this, that the probabilities of the two orders are the same. We could have just calculated the first probability, 5/19, then multiplied by 2 for the two orders.)
    Thanks from markosheehan
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  6. #6
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    Re: discs in a bag

    another way of doing it is using combinations.
    choose 2 from 20 = 20ncr2=190 total number

    (choose 1 from 20 even)and(chose 1 from 10 odd)
    10ncr1*10ncr1=100

    100/190=10/19

    i do not like this method as i am finding it hard to see why it works even though it gives the right answer
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