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Another probability question

  1. Mar 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Of the three independent events E1 , E2 and E3, the probability that only E1 occurs is α, only E2 occurs is β and only E3 occurs is γ. Let the probability p that none of the events E1 , E2 and E3 occurs satisfy the equations
    ## (α - 2β) p = αβ ## and ## (β - 3γ) p = 2βγ ##.
    All the given probabilities are assumed to lie in the interval (0,1).
    Then, (Probability of occurrence of E1) / (Probability of occurrence of E3) =

    Answer is 6.
    2. Relevant equations


    3. The attempt at a solution
    http://s3.amazonaws.com/minglebox-p...ata-0000-fdbffe7622c53ecd0122c5c50d0b0334.gif

    I don't know how to use those equations. All I know is what regions α, β, γ and p represent. α ≡ region 1, β ≡ region 3, γ ≡ region 7 and
    1 - regions(1+2+3+4+5+6+7) = p. How do I proceed?
     
    Last edited: Mar 21, 2016
  2. jcsd
  3. Mar 21, 2016 #2

    andrewkirk

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    The key is that the events are independent, so the probabilities of the joint occurrence or non-occurrence of different events is simply the product of the individual probabilities.

    To use that fact, define A, B, C to be the probabilities of occurrence of each of the three events. Then you can write ##\alpha,\beta,\gamma## and ##p## each in terms of A,B,C.

    Then your two equations above will be equations in terms of unknown probabilities A, B and C and you are asked to find A/C. See if you can rearrange the two equations so that they are in terms of the two unknowns B and A/C. If you can do that, you will have two equations and two unknowns, which you can solve.
     
  4. Mar 21, 2016 #3
    I did that actually, it didn't lead anywhere. Let me post it.
     
  5. Mar 21, 2016 #4

    andrewkirk

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    I tried it and it worked for me: I got 6. There's an awful lot of cancelling that happens when you write out the equations, and it simplifies nicely.
    Perhaps you made a misstep somewhere that stopped things from cancelling as they should.
     
  6. Mar 21, 2016 #5
    Let P(E1) = a , P(E2) = b and P(E3) = c.

    α = a - ab - ac + abc
    β = b - ab - bc + abc
    γ = c - ac - bc + abc
    p = (1-a)(1-b)(1-c)

    I can't use these to solve the equations. I tried various ways, it's not working.
     
  7. Mar 21, 2016 #6
    Oh perhaps. Let me check again.
     
  8. Mar 21, 2016 #7

    andrewkirk

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    Try the following. It'll be less messy, and hence less likely to generate mistakes. Write ##a'## for ##1-a, b'## for ##1-b, c'## for ##1-c##.
    Then you get ##\alpha=ab'c',\beta=a'bc',\gamma=ab'c',p=a'b'c'## which is much neater and easier to use.
    Write your two equations out using ##a,b,c,a',b',c'## and you'll find you can do a lot of cancelling and greatly simplify them before you need to convert ##a', b' c'## back to expressions in terms of ##a,b,c##.

    If it doesn't work, post what you did.
     
  9. Mar 21, 2016 #8
    Thank you so much!! I got it :)
     
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