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Math Help - Probability Question

  1. #1
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    Probability Question

    Hi, I have been stuck with this example for quite a while now. The question is as follows:



    A customer entering a bicycle store will buy a new bike with a probability of 10% and will buy bicycle accessories with a probability of 65%. With a probability of 30% a customer will not purchase anything.

    (a) Find the probability that a customer will buy both a new bike and bicycle accessories.
    and (b) Find the probability that a customer will only buy a bike.

    I have done similar questions before however with this question I don't quite understand what to do with.

    For part (a) I did the following:

    A = buy a bike. P(A) = 0.1
    B = buy bike accessories. P(B) = 0.65

    Then I tried to use the formula:
    P(A u B) = P(A) + P(B) - P(A n B) but I don't know how I would find P(A n B).

    In the end I got a value of 0.685 but don't think its correct.

    Thanks
    Last edited by axa121; September 29th 2010 at 03:04 PM.
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  2. #2
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    Quote Originally Posted by axa121 View Post
    Hi, I have been stuck with this example for quite a while now. The question is as follows:



    A customer enters a bicycle store and will buy a new bike with a probability of 10% and will buy bicycle accessories with a probability of 65%. With a probability of 30% that a customer will not buy anything.

    (a) Find the probability that a customer will buy a new bike and bicycle accessories.
    and (b) Find the probability that a customer will only buy a bike.

    I have done similar questions before however with this question I don't quite understand what to do with.

    For part (a) I did the following:

    A = buy a bike. P(A) = 0.1
    B = buy bike accessories. P(B) = 0.65

    Then I tried to use the formula:
    P(A u B) = P(A) + P(B) - P(A n B) but I don't know how I would find P(A n B).

    In the end I got a value of 0.685 but don't think its correct.

    Thanks
    Start by drawing a Karnaugh table that represents the given information:

    \begin{tabular}{l | c | c | c} & B & B' &  \\ \hline A &  &  & 0.1 \\ A' &  & 0.3 & \\ \hline &  0.65 &  & 1 \\ \end{tabular}<br />

    Fill in the gaps.

    Then get the required probability from the table.
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  3. #3
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    This is what I got for the table:


    A -------B-----------B'

    A-------0.05------0.05--------0.1

    A'-------0.6---------0.3-------0.9

    ---------0.65------0.35---------1
    Last edited by axa121; September 29th 2010 at 03:10 PM.
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  4. #4
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    Oh maybe I was interpreting it in a different way, as I was thinking that the B itself has a probability of 0.65.

    It's been a while since I've done stats, and the way I was reading it, I thought you would n't use the P(B|A) formula.

    I've used that now and got the value for P(B n A) as 0.065

    Is it correct?

    edit: I've also put the actual wording of the question up now.
    Last edited by axa121; September 29th 2010 at 03:11 PM.
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  5. #5
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    Quote Originally Posted by axa121 View Post
    Oh maybe I was interpreting it in a different way, as I was thinking that the B itself has a probability of 0.65.

    It's been a while since I've done stats, and the way I was reading it, I thought you would n't use the P(B|A) formula.

    I've used that now and got the value for P(B n A) as 0.065

    Is it correct? Mr F says: Is 0.065 the number in the cell of the Karnaugh table coresponding to \Pr(A \cap B)?

    edit: I've also put the actual wording of the question up now.
    ..
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  6. #6
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    No it is not.
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