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Math Help - Probability

  1. #1
    ttG
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    Probability

    Hi! Can somebody explain me please how to solve these:
    a) Calculate the probability of getTing at least one die showing a 5 using two dices?
    b) Calculate the prob. of getting the sum 10 and none of the dices are showing a 5?
    C) Calculate the prob. of getting two 5 if you may roll twice? That is if one of the dices doesn't show a 5 this one a second time and if none of the dices are showing a 5 you role both.
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by ttG View Post
    Hi! Can somebody explain me please how to solve these:
    a) Calculate the probability of getTing at least one die showing a 5 using two dices?
    b) Calculate the prob. of getting the sum 10 and none of the dices are showing a 5?
    C) Calculate the prob. of getting two 5 if you may roll twice? That is if one of the dices doesn't show a 5 this one a second time and if none of the dices are showing a 5 you role both.
    a) The question asks AT LEAST. So it does not rule out the possibility that you could throw TWO fives.

    The probability of throwing one 5 and any other number from 1 to 6 on the second die is: \frac{6}{36} = \frac{1}{6}

    b) Well in total we have 36 possibilities.

    So now let's see in how many ways can we make 10 without using a 5.

    We could get 6 and 4. And on another event we could get 4 and 6.

    So that's \frac{2}{36} = \frac{1}{18}

    c) I do not quite understand your question. But for this we need to use the binomial distribution. I do not know if you're familiar with it.

    The probability of getting 2 fives is \frac{1}{36}

    So according to our binomial distribution the probability is:

    {2 \choose 2} \left( \frac{1}{36} \right) ^{2} \left( \frac{5}{36} \right) ^{2-2} = \frac{5}{46656}
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  3. #3
    ttG
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    Smile

    My..concentration gone.. I've ment:
    Calculate the probability of getting two fives if you may roll twice? That is if one of the dices doesn't show a five you roll this one a second time and if none of this dices are showing a five you roll both!
    Thanks for the explanation!
    Last edited by ThePerfectHacker; June 2nd 2008 at 01:12 PM.
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by ttG View Post
    That is if one of the dices doesn't show a five you roll this one a second time and if none of this dices are showing a five you roll both!
    Here's a guess on it...

    It should follow the law of total probability i think.

    So we have three outcomes.

    1. Both are fives. (1/36)

    2. One is a five, now we roll again. (5/36 + 1/6)

    3. Neither is a five. Roll both again. (1/36)

    Add all these up.

    Answer: 13/36

    Not sure if this is correct though.
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  5. #5
    ttG
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    hmmm.. I think so! Well, if so..I have to evaluate one after another to get the sum of possible outcomes, right?
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by ttG View Post
    hmmm.. I think so! Well, if so..I have to evaluate one after another to get the sum of possible outcomes, right?
    I replied.
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  7. #7
    ttG
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    Thank u! It has to be right, at least i hope so
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  8. #8
    MHF Contributor arbolis's Avatar
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    I repost my post since the other thread has been closed.
    The probability of an event is equal to the number of favorable cases divided by the number of possible cases.
    For example I do your question a).

    a) Calculate the probability of getTing at least one die showing a 5 using two dices?
    You can make a table to show all the possible and favorable cases. A event will be a par like , or for example. As it is a 6x6 table, there are possible cases. And if you count the favorable cases, you realize that there are of them. So applying the simple formula I gave you, there is a probability of to get at least a 5 pitching once two dices.
    I've counted the case (5,5). I believe I'm right on the a), so the result wouldn't be \frac{13}{36} but \frac{11}{36}. I can easily enumerates the favorable cases : (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4) and (5,6). Therefore I'm almost sure it's \frac{11}{36}. I hope someone else can help us about this uncertain point.
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