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Math Help - Probability

  1. #1
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    Probability

    Find an example in which:
    P(AandB)>P(A)*P(B)

    P(AandB)<P(A)*P(B)
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  2. #2
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    Quote Originally Posted by ulysses123 View Post
    Find an example in which:
    P(AandB)>P(A)*P(B)

    P(AandB)<P(A)*P(B)
    Hint: A and B are not independent events.

    What have you tried? What thoughts have you had?
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  3. #3
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    Hello, ulysses123!

    Find an example in which: . \begin{array}{cccc}[1] & P(A \cap B)&>&P(A)\!\cdot\!P(B) \\ \\[-3mm] [2] &  P(A \cap B) &<& P(A)\!\cdot\!P(B)\end{array}

    We randomly select a number from: . \{2,3,4,5\}

    We have: . \begin{Bmatrix}P(\text{odd}) &=& \frac{2}{4} \\ \\[-3mm]<br />
P(\text{even}) &=& \frac{2}{4} \\ \\[-3mm]<br />
P(\text{prime}) &=& \frac{3}{4} \end{Bmatrix}


    P(\text{odd}\cap\text{prime}) \:=\:\frac{2}{4} \:=\:\frac{1}{2}

    P(\text{odd})\!\cdot\!P(\text{prime}) \:=\:\frac{2}{4}\cdot\frac{3}{4} \:=\:\frac{3}{8}

    Therefore: . [1]\;P(\text{odd}\cap\text{prime}) \;>\;P(\text{odd})\!\cdot\!P(\text{prime})


    P(\text{even}\cap\text{prime}) \:=\:\frac{1}{4}

    P(\text{even})\!\cdot\!P(\text{prime}) \:=\:\frac{2}{4}\!\cdot\!\frac{3}{4} \:=\:\frac{3}{8}

    Therefore: . [2]\;P(\text{even}\cap\text{prime}) \;<\;P(\text{even})\!\cdot\!P(\text{prime})

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  4. #4
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    what i have looked at is if the two events are dijoint, then the intersection is 0 and thus P(AandB)<P(A)*P(B)
    But can someone explain to me why two disjoint events are actually NOT independant.Thsi concept has been in the back of my mind since the start of semester, and i havn't really figured it out yet.
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  5. #5
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    Quote Originally Posted by ulysses123 View Post
    [snip]
    But can someone explain to me why two disjoint events are actually NOT independant.Thsi concept has been in the back of my mind since the start of semester, and i havn't really figured it out yet.
    Is \Pr(A \cap B) = \Pr(A) \cdot \Pr(B) true when A and B are disjoint?
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  6. #6
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    if they are disjoint the P(A and B)=0. Does this mean that the condition for independance can Never be fulfilled for disjoint events?
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  7. #7
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    If two sets are disjoint (and they have some positive probability), then they are DEPENDENT.
    This does make sense.
    Since IF you know A has occured then you know B cannot.
    Hence they do have a relationship.
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