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Math Help - Probability distribution table. Getting wrong values for X but can't find reason.

  1. #1
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    Probability distribution table. Getting wrong values for X but can't find reason.

    hello again.

    I have a problem with this question about a probability distribution table. I'm getting the wrong answer but I cannot figure why this is so. Here's the question:

    A small farm has 5 ducks and 2 geese. Four of these birds are to be chosen at random. The random variable X represents the number of geese chosen.

    (i) Draw up the probability distribution of X.

    I know that it will look like this:

    x 0 1 2
    P(X=x)

    I'm getting the values 0.26 for 0, 0.417 for 1 and 0.25 for 2. All of these are wrong, according to the book. Here's what I did:

    for P(X=0): 4C0(2/7)^0(5/7)^4 = 625/2401 or 0.2603
    for P(X=1): 4C1(2/7)^1(5/7)^3 = 1000/2401 or 0.4165
    for P(X=2): 4C2(2/7)^2(5/7)^2 = 600/2401 or 0.2498

    If I'm doing something wrong, I'd really like to know what it is because I've checked the above workings several times and keep getting the values above.

    Any help would be appreciated.
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  2. #2
    Senior Member Sambit's Avatar
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    The total number of ways in which 4 animals can be drawn from 7 animals is {{7}\choose{4}}.

    For X=0, all the 4 animals are drawn from the 5 ducks which can be done in {{5}\choose{4}} ways. So P(X=0)=\frac{{{5}\choose{4}}}{{{7}\choose{4}}}

    For X=1, 1 goose is chosen from 2 geese in {{2}\choose{1}} ways, the other 3 are chosen from the 5 ducks in {{5}\choose{3}} ways. So P(X=1)=\frac{{{2}\choose{1}}{{5}\choose{3}}}{{{7}\  choose{4}}}

    For X=2, 2 geese are taken from 2 geese in {{2}\choose{2}} ways, the remaining 2 being chosen from the 5 ducks in {{5}\choose{2}}. So P(X=2)=\frac{{{5}\choose{2}}{{2}\choose{2}}}{{{7}\  choose{4}}}

    The probabilities come out to be 0.142857, 0.571429 and 0.285714 respectively (which add up to 1, as expected).

    These answers has to be same as in your book.

    If you still have any confusion, you can ask.
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  3. #3
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    Thanks. Those are the correct values.

    Makes sense now. I've understood this, but if you are willing to answer, how would we have written all this in the form given in my first post? Just asking for knowing. If you don't reply, I've still understood the problem regardless.
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  4. #4
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    Your error was that you were treating this as "sampling with replacement". The probability of getting "4 ducks" is \left(\frac{5}{7}\right)^4 only if it is possible that the same duck can be chosen more than once. In "sampling without replacement", where once the duck is chosen, it cannot be chosen again, the probability of getting 4 ducks is (5/7)(4/6)(3/5)(2/4). Each time you choose a duck the number of ducks (the numerator) and the number of bird (the denominator) decreases.
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