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Math Help - box item problem

  1. #1
    Member
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    Apr 2010
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    dfsdfdf
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    Smile box item problem

    Dear Sir,
    I need help some help with the below problem and would appreciate very much if can offer some help.
    thanks
    Kingman

    John has 5 items, one each to be named P,Q, R,S and T. Each item is placed in a separate
    box and sealed. He then addresses the boxes, at random, one each to P,Q,R,S and T.
    1) Find the probability that the item P is in the correct box and the letter Q is in the
    wrong box. (3/20)
    2) Find the probability that the item P is in the correct box, given that the item Q is in the
    wrong box.(3/16)
    3) Find the probability that both of the items to P and Q are in the wrong boxes.(13/20)

    Hint: It would be helpful to draw a Venn diagram showing the various given probabilities.
    A suitable tree diagram is also helpful.
    Lastly let X denotes the event that Xs item is in the correct box.
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, kingman!

    I can't imagine how to draw a Venn diagram for this problem.


    John has 5 items named P,Q, R,S, T.
    Each item is placed in a separate box and sealed.
    He then addresses the boxes at random, one each to P, Q, R, S, T.

    1) Find the probability that the item P is in the correct box
    and item Q is in the wrong box. (3/20)
    First, probability that P is correct: . \frac{1}{5}
    Then probability that Q is incorrect: . \frac{3}{4}

    Therefore: . p\left(P \wedge \overline{Q}\right) \;=\;\frac{1}{5}\cdot\frac{3}{4} \;=\;{\color{blue}\frac{3}{20}}



    2) Find the probability that the item P is in the correct box,
    given that the item Q is in the wrong box. (3/16)
    Bayes' Theorem: . p\left(P\,|\,\overline{Q}\right) \;=\;\frac{p\left(P \wedge \overline{Q}\right)}{p\left(\overline{Q}\right)} .[1]

    We found the numerator in part (1): . p(P\,\wedge\,\overline{Q}) \;=\;\frac{3}{20}

    The denominator is: . p(\overline{Q}) \:=\:\frac{4}{5}

    Substitute into [1]: . p\left(P\,|\,\overline{Q}\right) \;=\; \frac{\frac{3}{20}}{\frac{4}{5}} \;=\;{\color{blue}\frac{3}{16}}




    3) Find the probability that both P and Q are in the wrong boxes. (13/20)
    This is a bit trickier . . .


    There are two scenarios:
    . . (1) P goes to Q.
    . . (2) P does not go to Q.


    (1) P goes to Q.

    . . .We have: . \_\;\;P\;\_\;\;\_\;\;\_

    . . The other four items can have any order
    . . and Q will be in the wrong box.

    . . .Hence: . p(P\to Q) \;=\;\boxed{\frac{1}{5}}


    (2) P does not go to Q.

    . . . . . . . . . . . . \overbrace{\qquad}^P
    . . . \text{We have: }\;\;\_\;\;\_\;\;\_\;\;\_\;\;\_

    . . . P can go to any of the last three addresses: . \frac{3}{5}
    . . .Then Q can go to three of the remaining four addresses: . \frac{3}{4}
    . . . Hence: . p(P \not\to Q) \;=\;\frac{3}{5}\cdot\frac{3}{4} \;=\;\boxed{\frac{9}{20}}


    Therefore: . p\left(\overline{P}\,\wedge\,\overline{Q}\right) \;=\;\frac{1}{5} + \frac{9}{20} \;=\;{\color{blue}\frac{13}{20}}

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  3. #3
    Member
    Joined
    Apr 2010
    From
    dfsdfdf
    Posts
    160

    thanks

    Dear Sir,

    Thanks very much for the neat and clear answer. However my main problem is how to express the outcomes,events and probabilities in the Venn diagrams or tree diagram. I wonder how I can express the solution in the form of P(Q|P),using intersection and union symbols.
    Thank you very much.
    Kingman
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