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Simple Probability Question (with p.d.f.)

  1. Apr 16, 2013 #1

    FeDeX_LaTeX

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    The problem statement, all variables and given/known data

    "The cakes in our canteen each contain exactly four currants, each currant being randomly placed in the cake. I take a proportion X of a cake where X is a random variable with density function

    f(x) = Ax

    for 0 ≤ x ≤ 1, where A is a constant.

    (i) What is the expected number of currants in my portion?
    (ii) If I find all four currants in my portion, what is the probability that I took more than
    half the cake?"

    The attempt at a solution

    First part seems pretty straightforward.

    [tex]\int_{0}^{1} Axdx = 1 \implies A = 2[/tex]

    So f(x) = 2x.

    [tex]E(X) = \int_{0}^{1} x \cdot 2xdx = \frac{2}{3}[/tex]

    The 4 currents are randomly distributed, so the expected number of currents in my portion is 4*(2/3) = 8/3.

    I can find the probability of taking more than half the cake (integrating the PDF between 1/2 and 1 gives 3/4), but not sure what to do about the 4 currants in my portion in part (ii). It's a conditional probability so I'd also need to know how to find the probability of having 4 currants AND more than half the portion, and I don't know how to do that... can anyone help me?
     
  2. jcsd
  3. Apr 16, 2013 #2

    Ray Vickson

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    For given proportion x, the probability that a current falls into it is x, so the distribution of the number of currents in it is binomial with parameters n = 4 and p = x. Can you take it from here?
     
  4. Apr 16, 2013 #3

    FeDeX_LaTeX

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    I see, so the probability of having 4 currants in my portion is x^4. But how can I use that and combine it with having more than half the cake?
     
  5. Apr 16, 2013 #4

    FeDeX_LaTeX

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    Wait, currants are independent of the proportion size, right? So can I just multiply x^4 by 2x and integrate? Or is that not the way...?
     
  6. Apr 16, 2013 #5

    haruspex

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    That would give you the probability of getting 4 currants, no?
    What equations do you know relating joint and conditional probabilities?
     
  7. Apr 17, 2013 #6

    FeDeX_LaTeX

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    Using P(A n B) = P(A|B)P(B), I'm getting 63/64. But I still can't intuitively understand why I can just multiply x^4 by 2x...
     
  8. Apr 17, 2013 #7

    Ray Vickson

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    You do not want the probability of an individual value of x; you want the probability that the proportion lies in the interval 1/2 <= x <= 1.
     
  9. Apr 17, 2013 #8

    FeDeX_LaTeX

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    So I can multiply them only because finding a current and the proportion size are independent?
     
  10. Apr 17, 2013 #9

    haruspex

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    They're not independent.
    Applying conditional probabilities here may be a little different from what you're used to because we have to consider 'individual' values in a continuous distribution. I.e. we work with the density function rather than actual probabilities of events. This can be made rigorous by considering the prob that the size of your slice is in some small range R(x) = (x, x+dx).
    P[currants = n & slice size in R(x)] = P[currants = n | slice size in R(x)] * P[slice size in R(x)]
    You should be able to write down the two terms on the RHS.
    Also
    P[currants = n & slice size in R(x)] = P[slice size in R(x) | currants = n] * P[currants = n]
     
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