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Probability of Independent Events

  1. Mar 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider three events A1, A2, and A3, and let pi = P(Ai), for i = 1, 2, 3.
    a) Express the probability that at least one of these three events occurs in terms of the pi ’s.
    b) Express the probability that at least two of the events occur.
    c) Suppose that A1, A2 and A3 are independent events. Verify that P(A1|A2 ∩ A3) = P(A1).

    2. Relevant equations


    3. The attempt at a solution
    For part c I had no trouble obtaining a solution:
    P(A1|A2 ∩ A3) = P[A1 ∩ (A2 ∩ A3)] P(A2 ∩ A3)
    By independence, we have P[A1 ∩ (A2 ∩ A3)] = P(A1)P(A2)P(A3) and P(A2 ∩ A3) = P(A2)P(A3), and the result follows. However, I am having some trouble with the first two. The answer given by the instructor for a is:
    1 − (1 − p1)(1 − p2)(1 − p3)
    and for b is:
    p1p2(1 − p3) + p1(1 − p2)p3 + (1 − p1)p2p3 + p1p2p3. I am not trying to dispute these answers, I am just having trouble understanding where they come from. Specifically, I do not understand the expression (1-p1), I assume that this is the probability that either p2 or p3 occur, but I'm not sure why. I was hoping somebody could give me some sort of explanation. Thanks.
     
  2. jcsd
  3. Mar 2, 2015 #2

    Orodruin

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    1-pi is the probability of Ai not occuring.
     
  4. Mar 2, 2015 #3
    Ok, thanks...so then the expression p1p2(1-p3) would be the probability that either p1 or p2 occur?
     
  5. Mar 2, 2015 #4

    Orodruin

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    It is the probability that A1 and A2 occur and that A3 does not - assuming the events are independent.
     
  6. Mar 2, 2015 #5
    ok, thank you
     
  7. Mar 2, 2015 #6

    Ray Vickson

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    The way the problem is written, independence does not enter until part (c), so (presumably), A1, A2, A3 need not be independent in parts (a) and (b). In that case you do not have enough information to do parts (a) and (b); you would need to know also the probabilities ##P(A_1 \cap A_2)##, ##P(A_1 \cap A_3##, ##P(A_2 \cap A_3## and ##P(A_1 \cap A_2 \cap A_3)##. If you did know these, the inclusion-exclusion principle would allow you to compute P{at least one }. A less well-known extension of the inclusion-exclusion principle would allow for calculation of P{exactly 1 occurs}, P{exactly 2 occur}, P{at least 2 occur}, etc. See, eg.,
    http://www.tricki.org/article/To_co...nts_occur_use_generalized_inclusion-exclusion

    If the events ARE independent, the inclusion-exclusion probabilities simplify a lot, and you can obtain explicit results. Below, let ##p_{i} = P(A_i)##, ##p_{ij} = P(A_i \cap A_j)## and ##p_{123} = P(A_1 \cap A_2 \cap A_3)## (with no independence assumptions).
    The instructor's answer for (a) comes from the fact that
    [tex] P(\text{at least one occurs}) = P(A_1 \cup A_2 \cup A_3)\\
    = p_1 + p_2 + p_2 - p_{12} - p_{13} - p_{23} + p_{123} .[/tex]
    Assuming indepedence, we can go further, as we would then have
    [tex] P(\text{at least one occurs})
    = p_1 + p_2 + p_3 - p_1 p_2 - p_1 p_3 - p_2 p_3 + p_1 p_2 p_3 \\
    = 1 - (1-p_1)(1 - p_2)(1 - p_3)
    [/tex]


    BTW: even "pairwise independence" is not enough: it is possible to have ##p_{12} = p_1 p_2##, ##p_{13} = p_1 p_3## and ##p_{23} = p_2 p_3##, but ##p_{123} \neq p_1 p_2 p_3##.
     
  8. Mar 2, 2015 #7
    Just to be clear, you simplification assuming independence comes from the fact that:
    [itex] P(A_{i1} \cap A_{i2} \cap ... \cap A_{ik}) = P(A_{i1}) * P(A_{i2}) *...*P(A_{ik}) [/itex] ?
     
  9. Mar 2, 2015 #8

    Ray Vickson

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    Yes.
     
  10. Mar 2, 2015 #9
    Ok fantastic, thank you for the explanation!
     
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