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Multiplication rule for conditional probabilities

  1. Sep 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Selecting Two Balls. Suppose that two balls are to be selected at random, without replacement, from a box containing r red balls and b blue balls. We shall determine the probability p that the first ball will be red and the second ball will be blue

    I am confusing Pr(A|B) and Pr(A∩B). In some cases they might look the same to me, so how can I recognize the difference between them.


    2. Relevant equations

    Pr(A∩B)=Pr(A) Pr(B|A)
    Pr(A∩B)=Pr(B) Pr(A|B).


    3. The attempt at a solution
    This is an example from the book, so the solution is already there.
    Let A be the event that the first ball is red,and let B be the event that the second ball is blue. Obviously
    , Pr(A)=r/(r +b)

    Furthermore, if the event A has occurred, then one red ball has been removed from the box on the first draw. Therefore, the probability of obtaining a blue ball on the second draw will be

    Pr(B|A)= b/( r +b−1)

    It follows that
    Pr(A∩B)= r/( r +b)* (b/( r +b−1))

    So the line " Furthermore, if the event A has occurred, then one red ball has been removed from the box on the first draw. Therefore, the probability of obtaining a blue ball on the second draw will be " looks the same to me as Pr(AnB) because the two events have to have occurred.

    .

    .
     
  2. jcsd
  3. Sep 9, 2015 #2

    Ray Vickson

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    No, you are confused: the joint probability ##P(A \cap B)## is the probability that BOTH ##A## AND ##B## will occur. The conditional probability ##P(B|A)## is the probability that ##B## will occur, GIVEN THAT ##A## is known to have occurred already.

    One of these refers to a probability of ##B## only (of course, in the new "sample space" ##A##!); the other refers to both occurring.
     
  4. Sep 12, 2015 #3

    haruspex

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    If you are still confused, it may help to consider A=B. P(B|B)=1, but ##P(B \cap B)=P(B)##.
     
  5. Sep 12, 2015 #4

    Ray Vickson

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    To expand on my initial answer, suppose r = b = 5, and A is an event which talks about what happened in the first 9 draws, while B = {last ball is blue}. What might be surprising is that P(B) = 1/2 (if you are told nothing at all about what happened in the first 9 draws---only that 9 balls have already been drawn and one is left in the urn). [More on this later.]

    Clearly, for some A we have P(B|A) = 0 (because there are no blue balls left), while for other A we have P(B|A) = 1 (because there is only a blue ball left). However, if A and B are not incompatible, then ##0 < P(A \cap B) < 1##; this will be the case for any A such that ##A \cap B \neq \emptyset##.

    Why ##P(B) = 1/2##? Number the balls from 1--10, and say balls 1--5 are blue. Look at all 10! permutations of the numbers 1--10, and look at those that have numbers 1--5 last. Exactly half of the permutations have that property, so ##P(B) = (1/2) 10! / 10! = 1/2##.
     
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