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    drfarah's Avatar
    drfarah Posts: 12, Reputation: 1
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    #1

    Oct 31, 2009, 09:50 AM
    Statistic (probability)
    I want to know the answer of this question URGENTLY please:

    Q:
    State the number of possible outcomes if a fair dice is drown two times. Write down all the possible outcomes that give the following events and calculate then the probability such that:
    a)The sum of the two tries is above 5

    b) The draws have the same outcome.

    Thanks a lot
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #2

    Oct 31, 2009, 11:30 AM
    There are 36 outcomes. List them out. 1,1; 1,2; 1,3; 1,4; 1,5; 1,6;... and so on.

    If there were 3 dice, then there would be 6^3 outcomes. If there were 10 dice, there would be 6^10 outcomes. See the pattern?

    There is a quick way to find how many outcomes sum to more than 5.

    That means they would have to sum to 6,7,8,9,10,11,12.

    It is called a generating function.



    If we expand this we get

    Look at the exponents and their corresponding coefficients(numbers in front of the x's).

    Look at the exponents larger than 5 and add up their coefficients. Each number it fron of the x is how many times that sum occurs. For instance, how many ways sum to 7? The coefficient corresponding to x^7 is 6. There are 6 ways the two dice sum to 7. So, the probability we get a sum of 7 is 6/36=1/6.
    See? Do this with your case and divide by 36 to fnd the probability. Or, you can list them out and count them up. But I do not like brute force. Especially with counting problems. It defeats the purpose.
    drfarah's Avatar
    drfarah Posts: 12, Reputation: 1
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    #3

    Nov 1, 2009, 04:01 AM

    So the pr(A), where A is the sm of the two tries above 5 = 27/36

    and the probability of the second Q is 6/36 (the probabilti of the two of tries are equal)

    Thank you very much for your patient and your helpful answers.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Nov 1, 2009, 05:02 AM

    I got 26/36!

    I used a probability table.
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #5

    Nov 1, 2009, 05:20 AM
    Yes, Unknown has it. Just add up the coefficients.

    They sum to 26

    So, we ave 26/36=13/18
    sajidbzu's Avatar
    sajidbzu Posts: 2, Reputation: 1
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    #6

    Nov 1, 2009, 11:40 AM

    what is the probability of getting one six if a die is thrown four times?
    sajidbzu's Avatar
    sajidbzu Posts: 2, Reputation: 1
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    #7

    Nov 1, 2009, 11:44 AM

    if a die is thrown four times what is the probability that the nunber is six?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #8

    Nov 1, 2009, 10:17 PM

    Start your own thread sakidbzu. And post the answer you think is good.
    drfarah's Avatar
    drfarah Posts: 12, Reputation: 1
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    #9

    Nov 2, 2009, 01:22 AM

    Oh OK thanks,
    And the draws have the same outcome will it be 6/36?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #10

    Nov 2, 2009, 01:31 AM

    Yes, exactly! :) Well done!
    drfarah's Avatar
    drfarah Posts: 12, Reputation: 1
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    #11

    Nov 2, 2009, 12:11 PM

    Thank you very much :)
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #12

    Nov 2, 2009, 08:26 PM

    You're welcome! :)

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