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Math Help - Probability using set notation

  1. #1
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    Probability using set notation

    The events A, B and C are such that A and B are independent, and A and C are mutually exclusive. Given that P(A)=0.4, P(B)=0.2, P(C)=0.3, P(B C)=0.1, calculate P(B/AC). Also, calculate the probability that one and only one of the events B, C will occur
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  2. #2
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    Lightbulb Re: Probability using set notation

    Hello, Punch!

    The events A, B, C are such that A and B are independent,
    and A and C are mutually exclusive.

    Given:. P(A)=0.4\;\;\;P(B)=0.2\;\;\;P(C)=0.3\;\;\;P(B \cap C)=0.1

    Draw a Venn diagram.
    Code:
                  * * *       * * *       * * *
              *     A     *     B     *     C     *
            *           *   *       *   *           *
           *           *     *     *     *           *
    
          *           *       *   *       *           *
          *   0.32    *  0.08 *   *  0.1  *    0.2    *
          *           *       *   *       *           *
    
           *           *     *     *     *           *
            *           *   *  0.02 *   *           *
              *           *           *           *
                  * * *       * * *       * * *
    A and C are mutually exclusive.
    . . Their circles do not intersect.

    A and B are independent.
    Hence:. P(A \cap B) \:=\:P(A)\cdot P(B) \:=\:(0.4)(0.2) \:=\:0.08

    P(B \cap C) \,=\,0.1

    We can complete the Venn diagram.




    (a) Calculate P(B\,|\,A \cup C)

    Bayes' Theorem: . P(B\,|\,A\cup C) \;=\;\frac{P(B \cap [A\cup C])}{P(A\cup C)}


    From the Venn diagram, we see that:
    . . P(B\cap[A \cup C]) \:=\:0.08 + 0.1 \:=\:0.18
    . . P(A \cup C) \:=\:0.4 + 0.3 \:=\:0.7

    \text{Therefore: }\:P(B\,|\,A\cup C) \:=\:\frac{0.18}{0.70} \:=\:\frac{9}{35}




    (b) Calculate the probability that exactly one one of B, C will occur.

    From the Venn diagram, we see that:
    . . P(B\text{ only}) \:=\:0.1
    . . P(C\text{ only}) \:=\:0.2

    Therefore:. P(B\text{ only}\,\cup\, C\text{ only}) \:=\:0.1 + 0.2 \:=\:0.3

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  3. #3
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    Re: Probability using set notation

    Thank you very much! Venn diagrams are certain useful in solving the question, but I have a doubt:

    Based on the rule for independent events, P(A\capB)=P(A)P(B)

    So, why is this not true: P(B|A\cupC)=\frac{P(B\cap[A\cup C])}{P(A\cup C)}=\frac{P(B)P(A\cup C)}{P(A\cup C)}=P(B)
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