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Math Help - probability help plz

  1. #1
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    probability help plz

    Hi there,
    I am doing s1 for this jan and i am finding it very difficult to cope up. Especially for probability. I have a cgp buk but stil its not very gud at probability. Here is a question from my text buk which i cud not understand : -



    Two events A and B are such that P(A) = 3/4 , P(B l A) = 1/5 and P(B' l A') = 4/7. By use of a probability tree or other wise find :


    a) P(A and B)

    b) P(B)

    c) P(A l B)


    In addition, can some 1 suggest a gud buk for probability plz.


    Thanks!
    Last edited by hasnain721; December 22nd 2008 at 12:15 PM.
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  2. #2
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    Hello, Hasnain!

    Two events A and B are such that: . P(A) \:=\:\frac{3}{4},\;\;P(B|A)\:=\:\frac{1}{5},\;\;P(  B'|A') \:= \:\frac{4}{7}

    Find: . (a)\;P(A \cap B)\qquad (b)\;P(B)\qquad (c)\;P(A|B)
    I assume you're familiar with Bayes' Theorem: . P(X|Y) \:=\:\frac{P(X \cap Y)}{P(Y)}

    (a) We have: . P(B|A) \:=\:\frac{P(B \cap A)}{P(A)}

    Since P(B|A) = \frac{1}{5} and P(A) = \frac{3}{4}

    . . the equation becomes: . \frac{1}{5} \:=\:\frac{P(A \cap B)}{\frac{3}{4}}\quad\Rightarrow\quad P(A \cap B) \:=\:\frac{1}{5}\cdot\frac{3}{4}

    Therefore: . \boxed{P(A \cap B) \;=\;\frac{3}{20}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    (b) There are a number of ways to determine P(B).
    Here's one of them . . .

    We are given: . P(B'|A') \:=\:\frac{P(B' \cap A')}{P(A')} \:=\:\frac{4}{7}

    Since P(A) = \frac{3}{4}, then P(A') = \frac{1}{4}
    . . and the equation becomes: . \frac{P(A' \cap B')}{\frac{1}{4}} \:=\:\frac{4}{7}\quad\Rightarrow\quad P(A' \cap B') \:=\:\frac{1}{7}


    Sketching a Venn diagram, we see that:

    . . P(A) + P(A \cap B) + P(A' \cap B') + P(A' \cap B) \;=\;1

    . . . \frac{3}{5}\quad + \quad\; \frac{3}{20}\quad \;\;+\quad\quad \frac{1}{7} \quad+\quad P(A'\cap B) \;=\;1

    Then: . P(A' \cap B) \:=\:1 - \frac{3}{5} -\frac{3}{20} - \frac{1}{7}\;=\;\frac{3}{28}

    Hence: . P(B) \;=\;P(A\cap B) + P(A' \cap B) \;=\;\frac{3}{20} + \frac{3}{28}

    Therefore: . \boxed{P(B)\;=\;\frac{9}{35}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    (c) We've done the hard work ... This part is easy . . .

    P(A|B) \;=\;\frac{P(A\cap B)}{P(B)}\;=\;\frac{\dfrac{3}{20}}{\dfrac{9}{35}} \;=\;\boxed{\frac{7}{12}}

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  3. #3
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    In addition, can some 1 suggest a gud buk for probability plz.
    I quite like Gnedenko,s "Theory of Probability".
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  4. #4
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    Dec 2007
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    Hi there,
    Thanks a lot for replying. I understand what u've done till you worked out
    P( A' ∩ B'). Is that the same as P( B' ∩ A') ?


    Secondly, when you said, " Sketching a Venn diagram, we see that:",
    i dont understand which part of the diagram is P(A' ∩ B) . If you dont mind, can u sketch it plz?



    Thanks a lot for replyin and i have checked the answers at the back of the book and they are right!


    Thanks.
    Hasnain Mir Mohammed
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