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Thread: How was this probability distribution calculated?

  1. #1
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    How was this probability distribution calculated?

    I'm reading my textbook, and I don't follow how the following probability distribution was calculated.

    How was this probability distribution calculated?-image.png


    Can you help me understand how we got the f(x) values in the probability distribution? What formulas were used? Is it some kind of counting calculation?
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  2. #2
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    Re: How was this probability distribution calculated?

    Quote Originally Posted by Fischercr View Post
    I'm reading my textbook, and I don't follow how the following probability distribution was calculated.

    Click image for larger version. 

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    Can you help me understand how we got the f(x) values in the probability distribution? What formulas were used? Is it some kind of counting calculation?
    $f(k)=\dbinom{3}{k}\left( \dfrac{1}{4}\right)^k\left( \dfrac{3}{4}\right)^{3-k}$___$k=0,1,2,3$

    $\dbinom{3}{k}=\left( \dfrac{3!}{(k!)(3-k)!}\right)$
    Last edited by Plato; Oct 8th 2017 at 08:57 AM.
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  3. #3
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    Re: How was this probability distribution calculated?

    Equivalently, we are told that "this process produces 25% defectives" so for any one item the probability it is defective (D) is 0.25= 1/4, the probability it is non-defective (N) is 075= 3/4.
    The probability of NNN, all three are non-defective, since we are also told these are "independent", is (3/4)(3/4)(3/4)= 27/64. With X the number of defective items, P(X= 0)= 27/64.

    The probability the first chosen is defective, the other two are non-defective, NND, in that order, is (3/4)(3/4)(1/4)= 9/64, But it is easy to see that the probability of "non-defective, defective, non-defective", NSN, in [b]that[/b ] order is (3/4)(1/4)(0.75)= 9/64 also and similarly "non-defective, non-defective, defective", NND, is also 9/64. P(X= 1)= 9/64+ 9/64+ 9/64= 3(9/64)= 27/64.

    The probability the both the first two chosen are defective, the third non-defective, DDN, in that order, is (1/4)(1/4)(3/4)= 3/64. Again, it is easy to see that the probabililties of "defective, non-defective, defective", DND, and "non-defective, defective, defective", NDD are (1/4)(3/4)(1/4)= 3/64, and of (3/4)(1/4)(1/4)= 3/64 also. P(X= 2)= 3/64+ 3/64+ 3/64= 3(3/64)= 9/64.

    Finally, the probability of "DDD", all three defective, is (1/4)(1/4)(1/4)= 1/64. P(X= 3)= 1/64.
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