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Math Help - HELP probability homework

  1. #1
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    Exclamation HELP probability homework

    Chris is going to roll a biased dice.
    The probabilty that he will get a six is 0.09.
    Chris is going to roll the dice 30 times.
    Work out an estimate for the number of sixes he will get...
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  2. #2
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    Quote Originally Posted by Nat077 View Post
    Chris is going to roll a biased dice.
    The probabilty that he will get a six is 0.09.
    Chris is going to roll the dice 30 times.
    Work out an estimate for the number of sixes he will get...
    The number of 6's rolled follows a binomial distribution.
    n = 30, p = 0.09.
    Just use the standard formula for expected value: np(1-p).

    Edit: This formula gives variance, NOT expected value. Thanks Nyoxis - good get.
    Last edited by mr fantastic; March 22nd 2008 at 05:32 AM.
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    answer

    Sorry i am new today i dont really understand what i am doing to be honest i thought about the advice you gave me and for my answer should i just put 2?
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  4. #4
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    Quote Originally Posted by Nat077 View Post
    Sorry i am new today i dont really understand what i am doing to be honest i thought about the advice you gave me and for my answer should i just put 2?
    For the benefit of all forum members, I'll quote my pm reply:

    Quote Originally Posted by Mr Fantastic in a pm
    Good question. The expected value is 2.457. If you did the 30 roll thing a large numbers of times, you'd expect to see the average be around 2.457 .....

    Doing it once, I'd be inclined to first give my answer as 2.457. Then I'd round to 3 on the grounds that you want a whole number and it's gotta be larger than 2.457 .....

    But ..... it might be that your instructor is looking for 2 as the answer ..... Best to see what he/she says, perhaps.
    [snip]
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  5. #5
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    Question

    Quote Originally Posted by mr fantastic View Post
    The number of 6's rolled follows a binomial distribution.
    n = 30, p = 0.09.
    Just use the standard formula for expected value: np(1-p).
    I thought E(X)=np, and Var(X)=np(1-p), so he needs to use E(X) here being...

    E(X)=np= 30x0.09= 2.7

    ...rounded to 3.

    Don't take this as an answer till someone else confirms it though, I'm tired today. =)
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  6. #6
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    Quote Originally Posted by Nyoxis View Post
    I thought E(X)=np, and Var(X)=np(1-p), so he needs to use E(X) here being...

    E(X)=np= 30x0.09= 2.7

    ...rounded to 3.

    Don't take this as an answer till someone else confirms it though, I'm tired today. =)
    Thank you. Great save.

    The expected value IS np. Your calculation and rounding is correct.

    I'm off to bed!
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