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Math Help - probability distribution

  1. #1
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    probability distribution

    how would i prepare a probability distribution of the X variable for this question?

    You have a biased coin or unfair coin, such that, on each toss, the probability of observing a head is 0.25 and the probability of observing a tail is 0.75. Let X be the variable = number of heads observed in 4 tosses.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by skhan
    how would i prepare a probability distribution of the X variable for this question?

    You have a biased coin or unfair coin, such that, on each toss, the probability of observing a head is 0.25 and the probability of observing a tail is 0.75. Let X be the variable = number of heads observed in 4 tosses.
    The possible outcomes are:

    hhhh p=0.25^4

    thhh p=0.25^3 0.75
    hthh p=0.25^3 0.75
    hhth p=0.25^3 0.75
    hhht p=0.25^3 0.75


    tthh p=0.25^2 0.75^2
    thth p=0.25^2 0.75^2
    thht p=0.25^2 0.75^2
    htth p=0.25^2 0.75^2
    htht p=0.25^2 0.75^2
    hhtt p=0.25^2 0.75^2

    ttth p=0.25 0.75^3
    ttht p=0.25 0.75^3
    thtt p=0.25 0.75^3
    httt p=0.25 0.75^3


    tttt p=0.75^4

    So

    P(0 heads)=p(tttt)=0.75^4
    P(1 heads)=4 p(httt)=4 0.25 0.75^3
    P(2 heads)=6 p(hhtt)=6 0.25^2 0.75^2
    P(3 heads)=4 p(hhht)=4 0.25^3 0.75
    P(4 heads)=p(hhhh)=0.25^4

    This can also be done using the formulae for the binomial distribution.

    RonL
    Last edited by CaptainBlack; May 12th 2006 at 10:13 PM. Reason: correction of typos
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  3. #3
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    thanks Ron

    from the probabilities that you gave me, i tried to compute the average however i am getting an answer of avg = 1.14 whereas the right answer is given as avg = 1.0

    can you spot where i am going wrong?

    P(0 heads)=p(tttt)=0.75^4 = 0.31640
    P(1 heads)=4 p(httt)=4 0.25 0.75^3 = 0.4218
    P(2 heads)=6 p(hhtt)=6 0.25^2 0.75^2 = 0.21093
    P(3 heads)=4 p(hhht)=6 0.25^3 0.75 = 0.07031
    P(4 heads)=p(hhhh)=6 0.25^4 = 0.0234375

    avg = 0(.31640) + 1(0.4218) + 2(0.21093) + 3(0.07031) + 4(0.02343)
    = 0.4218 + 0.42186 + 0.21093 + 0.09372
    = 1.14
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by skhan
    thanks Ron

    from the probabilities that you gave me, i tried to compute the average however i am getting an answer of avg = 1.14 whereas the right answer is given as avg = 1.0

    can you spot where i am going wrong?

    P(0 heads)=p(tttt)=0.75^4 = 0.31640
    P(1 heads)=4 p(httt)=4 0.25 0.75^3 = 0.4218
    P(2 heads)=6 p(hhtt)=6 0.25^2 0.75^2 = 0.21093
    P(3 heads)=4 p(hhht)=6 0.25^3 0.75 = 0.07031
    P(4 heads)=p(hhhh)=6 0.25^4 = 0.0234375

    avg = 0(.31640) + 1(0.4218) + 2(0.21093) + 3(0.07031) + 4(0.02343)
    = 0.4218 + 0.42186 + 0.21093 + 0.09372
    = 1.14
    I'm afraid that is because some typos crept in during the cut and past job
    used to compite the table, it should read:

    P(0 heads)=p(tttt)=0.75^4
    P(1 heads)=4 p(httt)=4 0.25 0.75^3
    P(2 heads)=6 p(hhtt)=6 0.25^2 0.75^2
    P(3 heads)=4 p(hhht)=4 0.25^3 0.75
    P(4 heads)=p(hhhh)=1 0.25^4

    which you should be able to see from the proceeding explanation of how
    it was derived.

    RonL
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