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Math Help - Solving Probability of Bernoulli Urn with the Expectation of the Fraction of Balls...

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    Solving Probability of Bernoulli Urn with the Expectation of the Fraction of Balls...

    Hello, I’m an undergraduate medical student (i.e, no mathematics education past high school), and I’m having difficulty understanding a concept in ET Jayne’s Probability Theory: The Logic of Science. On pp.63-67 2ed., he discusses a Bernoulli urn with N = 4 balls, M = 2 red ones (and 2 white, N-M), of which we must randomly draw n= 3. The balls are not replaced. (Let this proposition ≣ B.) He asks, how does knowledge that a red ball will be drawn on the second (R2) or third (R3) draw affect the probability of drawing a red ball on the first (R1)?


    He reveals the surprising (and awesome!) revelation that P(R1 | R2 + R3,B) > P(R1 | R2 B). I understand his intuitive explanation for it, but not his formal. He summarises thusly:


    “... when the fraction F = M/N of red balls is known, then the Bernoulli urn rule applies, and P(R1 | B) = F. When F is unknown, the probability for red is the expectation of F: P(R1 | B) = <F> ≣ E(F). If M and N are both unknown, the expectation is over the joint probability distribution for M and N.”


    I tried calculating E(F), (as I assume the second and not the third scenario applies here), but arrived to an erroneous result. In the intuitive working, he shows P(R1 | R2 + R3,B) = (4/5) / 2, calling the numerator ‘effective M’ and the denominator is ‘N - 2’. In his formal explanation, he states that ‘effective M’ is the Expected value of M, E(M).


    So, I don’t know have that fits into finding the Expected value of F. Basically, I don’t know where the N term fits into it all. Any help would be much appreciated; if I have been too unclear I will make screenshots of the pages.
    Last edited by Gazmann; December 31st 2012 at 10:52 PM.
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    Re: Solving Probability of Bernoulli Urn with the Expectation of the Fraction of Ball

    Quote Originally Posted by Gazmann View Post
    Hello, I’m an undergraduate medical student (i.e, no mathematics education past high school), and I’m having difficulty understanding a concept in ET Jayne’s Probability Theory: The Logic of Science. On pp.63-67 2ed., he discusses a Bernoulli urn with N = 4 balls, M = 2 red ones (and 2 white, N-M), of which we must randomly draw n= 3. The balls are not replaced. (Let this proposition ≣ B.) He asks, how does knowledge that a red ball will be drawn on the second (R2) or third (R3) draw affect the probability of drawing a red ball on the first (R1)?


    He reveals the surprising (and awesome!) revelation that P(R1 | R2 + R3,B) > P(R1 | R2 B). I understand his intuitive explanation for it, but not his formal. He summarises thusly:


    “... when the fraction F = M/N of red balls is known, then the Bernoulli urn rule applies, and P(R1 | B) = F. When F is unknown, the probability for red is the expectation of F: P(R1 | B) = <F> ≣ E(F). If M and N are both unknown, the expectation is over the joint probability distribution for M and N.”


    I tried calculating E(F), (as I assume the second and not the third scenario applies here), but arrived to an erroneous result. In the intuitive working, he shows P(R1 | R2 + R3,B) = (4/5) / 2, calling the numerator ‘effective M’ and the denominator is ‘N - 2’. In his formal explanation, he states that ‘effective M’ is the Expected value of M, E(M).


    So, I don’t know have that fits into finding the Expected value of F. Basically, I don’t know where the N term fits into it all. Any help would be much appreciated; if I have been too unclear I will make screenshots of the pages.

    Frankly I do not follow the text. But the table below gives all possible outcomes. We are looking for \mathcal{P}(R_1|R_2\cup R_3).

    \begin{array}{*{20}c}   I &\vline &  {II} &\vline &  {III}  \\\hline   R &\vline &  R &\vline &  W  \\   R &\vline &  W &\vline &  R  \\   R &\vline &  W &\vline &  W  \\   W &\vline &  R &\vline &  R  \\   W &\vline &  R &\vline &  W  \\   W &\vline &  W &\vline &  R  \\\end{array}

    Of those six we concentrate on the rows which have an R in the second or third column. That is the given part. There are only five of those.
    Of those five, only two also have an R in the first column.

    Thus \mathcal{P}(R_1|R_2\cup R_3)=\frac{2}{5}~.
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    Re: Solving Probability of Bernoulli Urn with the Expectation of the Fraction of Ball

    Thank you Plato, that part I understand

    Here is the section in question, imgur: the simple image sharer .

    Referring to the text, my confusion lies with reconciling Equation 3.63 with the paragraph below Equation 3.71 (explaining Eq.3.63 in a more 'cogent' way...).
    It mostly lies in the fact I don't understand how to find the Expectation of the Fraction, M/N. (I understand finding the Expectation of M, but not when N is involved.)
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