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Math Help - Multiple event probability

  1. #1
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    Exclamation Multiple event probability

    Hi, for some reason I'm having issues with something I usually find easy. The question is as follows:

    events A and B are such that P(A) = 2/5, P(B)= 1/2 and P(A|B') = 4/5

    Find

    a.) P(A n B')
    b.) P(A n B)
    c.) P(A u B)
    d.) P(A|B)

    next, state with reason, whether or not A and B are

    a) Mutually exclusive
    b) independent

    What I've done:

    When I try and draw out a Venn diagram (As i find them helpful for displaying data) can get P(A) to equal 4/5 when given B' however when I try and work out P(A n B) for the middle of the Venn diagram it comes out as 0.2, when subtracting that from the probabilities of A and B I end up with 0.2 for A. Which no matter what I do I can't make P(A|B') equal 4/5. So I can make my Venn diagram work and I have no idea whether to use the values without subtracting (A n B) or not. I was wondering if they were independent and there were no cases where they are both chosen so P(A n B) would = 0?

    Please help it would be much appreciated. I look forward to hearing from you!

    BigGScotty278 (AS level Maths & Further Maths)
    Last edited by BigGScotty278; January 3rd 2014 at 06:41 AM.
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  2. #2
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    Re: Multiple event probability

    Quote Originally Posted by BigGScotty278 View Post
    Hi, for some reason I'm having issues with something I usually find easy. The question is as follows:

    events A and B are such that P(A) = 2/5, P(B)= 1/2 and P(A|B') = 4/5

    Find

    a.) P(A n B')
    b.) P(A n B)
    c.) P(A u B)
    d.) P(A|B)

    next, state with reason, whether or not A and B are

    a) Mutually exclusive
    b) independent

    What I've done:

    When I try and draw out a Venn diagram (As i find them helpful for displaying data) can get P(A) to equal 4/5 when given B' however when I try and work out P(A n B) for the middle of the Venn diagram it comes out as 0.2, when subtracting that from the probabilities of A and B I end up with 0.2 for A. Which no matter what I do I can't make P(A|B') equal 4/5. So I can make my Venn diagram work and I have no idea whether to use the values without subtracting (A n B) or not. I was wondering if they were independent and there were no cases where they are both chosen so P(A n B) would = 0?

    Please help it would be much appreciated. I look forward to hearing from you!

    BigGScotty278 (AS level Maths & Further Maths)

    P(A)=P(A|B)P(B)+P(A|B')P(B')

    \frac{2}{5}=P(A|B) \frac{1}{2}+\frac{4}{5} \left(1-P(B)\right)=P(A|B)\frac{1}{2}+\frac{4}{5} \cdot \frac{1}{2}

    \frac{2}{5}=P(A|B)\frac{1}{2}+\frac{2}{5}

    \frac{1}{2}P(A|B)=\frac{2}{5}-\frac{2}{5}=0 so

    P(A|B)=0

    see if that lets you make headway.
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  3. #3
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    Re: Multiple event probability

    Hello, BigGScotty278!

    \text{Given: }\:P(A)\,=\,0.4,\;P(B)\,=\,0.5,\;P(A|B')\,=\,0.8

    \text{Find: }\;(a)\;P(A \cap B') \qquad (b)\;P(A \cap B) \qquad (c)\;P(A \cup B) \qquad (d)\;P(A|B)

    \begin{array}{ccccc}\text{We have:} & P(A) \:=\:0.4 & P(A') \:=\:0.6 \\ & P(B) \:=\:0.5 & P(B') \:=\:0.5 \end{array}

    P(A|B') \:=\:0.8 \quad\Rightarrow\quad \frac{P(A\cap B')}{P(B')} \:=\:0.8 \quad\Rightarrow\quad P(A\cap B') \:=\:(0.8)\!\cdot\!P(B')

    . . P(A \cap B') \:=\:(0.8)(0.5) \quad\Rightarrow\quad P(A\cap B') \:=\:0.4


    With the above information, we have this chart:

    . . \begin{array}{c||c|c||c|} & B & B' & \text{Total} \\ \hline\hline A && 0.4 & 0.4 \\ \hline A' & && 0.6 \\ \hline\hline \text{Total} & 0.5 & 0.5 & 1.0 \\ \hline \end{array}


    We can complete the chart:

    . . \begin{array}{c||c|c||c|} & B & B' & \text{Total} \\ \hline\hline A &0.0& 0.4 & 0.4 \\ \hline A' & 0.5 & 0.1 & 0.6 \\ \hline\hline \text{Total} & 0.5 & 0.5 & 1.0 \\ \hline \end{array}


    And answer the questions:

    . . (a)\;P(A\cap B') \:=\:0.4

    . . (b)\;P(A \cap B) \:=\:0

    . . (c)\;P(A\cup B) \:=\:0.9

    . . (d)\;P(A|B) \:=\:\frac{P(A\cap B)}{P(B)} \:=\:\frac{0}{0.5} \:=\:0




    \text{Next, state with reason, whether or not }A\text{ and }B\text{ are:}

    . . \text{(a) Mutually exclusive} \qquad \text{(b) independent}

    \text{(a) Since }P(A\cap B) = 0,\:A\text{ and }B\:are\text{ mutually exclusive.}

    (b) \;P(A)\!\cdot\!P(B) \:=\:(0.4)(0.5) \:=\:0.2
    . . P(A \cap B) \:=\:0

    \text{Since }P(A)\!\cdot\!P(B) \,\ne\,P(A\cap B),\,A\text{ and }B\text{ are }not\text{ independent.}
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