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  • Set Theory: Show that P(A U B) = P(A) + P(Ac ∩ B)?

    Anonymous Christian
    Sep 5, 2014
    MATH
    Comments Off on Set Theory: Show that P(A U B) = P(A) + P(Ac ∩ B)?

    Answer

    P(A U B) = P(A) + P(Ac ∩ B)?

    Here is a simple but understandable approach. The method below is called the “counting argument” in combinatorics and probability. Since we have “two events” A & B, lets define the number of elements as follows :

    A has ‘a’ elements. We write A = a for simplicity. Similarly,

    A’ = a’ ( Ac, the number of elements in the complement of A )
    B = b
    B’ = b’
    A n B = c
    Total elements = y

    Now A U B = a + b – c
    So, P ( A U B ) = (No. of element in A U B )/ ( Total elements )
    = ( a + b – c ) / y …(1)

    Similarly, we establish that :
    P(A) = a/y …(2) and

    Consider (Ac n B). It contains the elements of B together with those that are in A’ or Ac. So, this is just the elements in B minus the elements that belong to both A and B. ( Draw a diagram or do an example and you will see this ),That is,
    (Ac n B) = B – ( A n B ) = b – c, Thus,
    P(Ac n B) = (b – c)/y…(3)

    “Combine” (1), (2) and (3) to get,

    P ( A U B ) = ( a + b – c ) / y
    = (a/y) + (b -c)/y
    = P(A) + P(Ac n B)

    QED

anonymous-christian

Jonathan Ramachandran is founder at AnonymousChristian.org which serves as a Christian Blog discussing the Greater Hope of Biblical Christian in Christ Alone. He is also an IMO (International Math Coach) and a featuring Guitarist for many bands.

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