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Math Help - Probability for exactly one of two events

  1. #1
    Member oldguynewstudent's Avatar
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    Probability for exactly one of two events

    A utility company offers a lifeline rate to any household whose electricity usage falls below 240 kWh during a particular month. Let A denote the event that a randomly selected household in a certain community does not exceed the lifeline usage during January, and let B be the analogous event for the month of July (A and B refer to the same household). Suppose P(A) = .8, P(B) = .7, and P(A\cup B)=.9. Compute the following:

    a) P(A\cap B) . P(A\cup B)=P(A)+P(B)-P(A\cap B). So P(A\cap B)= .8 + .7 - .9 = .6

    b) The probability that the lifeline usage amount is exceeded in exactly one of the two months. Describe this event in terms of A and B. I believe this would be an XOR relationship between A and B, so that would be P(A\cup B)-P(A\cap B) = .9 - .6 = .3.

    Please let me know if these calculations and set relationships are accurate.
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  2. #2
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    As you said this is an Xor.
    You want P\left( {A \cap \overline B } \right) + P\left( {\overline A  \cap B} \right), where \overline B is not B.

    Notice that P\left( {A \cap \overline B } \right) = P(A) - P(A \cap B).
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  3. #3
    Member oldguynewstudent's Avatar
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    Lightbulb

    Quote Originally Posted by Plato View Post
    As you said this is an Xor.
    You want P\left( {A \cap \overline B } \right) + P\left( {\overline A  \cap B} \right), where \overline B is not B.

    Notice that P\left( {A \cap \overline B } \right) = P(A) - P(A \cap B).
    Thank you, so then it follows that  P\left( {\overline A  \cap B} \right) = P(B) - P(A \cap B).

    Which when you take P\left( {A \cap \overline B } \right) + P\left( {\overline A  \cap B} \right) = P(A) + P(B) - P(A \cap B) - P(A \cap B). But taking away the intersection twice is equivalent to taking it away once since the second subtraction would already be empty. Correct?\

    I see my error, we are dealing with the probabilities so you would subtract the intersection twice!
    Last edited by oldguynewstudent; June 19th 2010 at 12:28 PM. Reason: incorrect logic
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    P\left( {A \cap \overline B } \right) + P\left( {\overline A  \cap B} \right) = P(A) + P(B) - 2P(A \cap B)
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  5. #5
    Member oldguynewstudent's Avatar
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    Red face

    Yes, thanks, I was editting my reply but you drew and shot first.
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  6. #6
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    That is the standard counting rule for the XOR.
    You will use it for the symmetric difference operator.
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