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Probability of Credit Cards P(A|B')

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  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=51608&stc=1&d=1349558719.jpg

    2. Relevant equations



    3. The attempt at a solution

    [tex]P(A | M^{'}) = \frac{P(A \cap M^{'})}{P(M^{'})}[/tex]

    I know that,

    [tex]P(M^{'}) = 1-0.48 = 0.52[/tex]

    but I can't figure out how to obtain,

    [tex]P(A \cap M^{'})[/tex]

    Any ideas?
     

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  2. jcsd
  3. Oct 6, 2012 #2

    Simon Bridge

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    Isn't it given to you? Probability has a mastercard and an amex?
     
  4. Oct 6, 2012 #3

    Dick

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    No, it's probability of NOT having a mastercard and having an amex. I'd suggest to jegues to draw a Venn diagram and start labelling regions.
     
  5. Oct 7, 2012 #4

    Simon Bridge

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    Oh I see what you mean - did you construct the ven diagram?

    ##A## = has an amex
    ##M## = has a mastercard
    ##A\cap M## has a mastercard and an amex.

    But assign an arbitrary number of people ... say 100.

    48 have a mastercard (52 don't)
    20 have an american express
    11 have a mastercard and an amex

    Thus, 37 people with a mastercard do not have an amex.
    So the probability that someone with a mastercard does not have an amex is 37/48.

    That's probably the quickest way to understand the problem ... you can rework it all in terms of conditional probabilities later. (To complete this diagram: how many people have neither an amex nor a master-card?)
     
  6. Oct 7, 2012 #5
    See figure attached for my attempt at drawing a Venn diagram.

    I'm having some trouble finding the probability of not having a mastercard or a visa or a amex.(i.e. being outside of all 3 circles)

    My attempt at this would be,

    [tex]0.7 + (0.48-0.29) + (0.2 - (0.11 + 0.14 - 0.06)) = 0.9[/tex]

    but that answer doesn't make any sense.

    Can someone explain how to do this? Do I need to do this to get closer to finding my answer?

    EDIT: After thinking about it more,

    [tex]P(A \cap M^{'}) = 0.2-0.11 = 0.09 \quad \quad P(M^{'}) = 0.52[/tex]

    [tex]P(A|M^{'}) = \frac{0.09}{0.52} = 0.17307[/tex]
     

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    Last edited: Oct 7, 2012
  7. Oct 7, 2012 #6

    Simon Bridge

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    You only need two circles like I said: see post #4.
    I think that's as close as I can get to telling you how to do it without doing the problem for you.

    If there are 100 people all together:
    1. How many people, in total, have an amex? (given)
    2. How many of the people who have an amex also have a mastercard? (given)
    3. How many people who have an amex don't have a mastercard? (from 1 and 2)
    4. What is the total number of people without a mastercard?
    5. what is the probability that someone without a mastercard has an amex? (from 3 and 4)

    That should help you understand what is behind the equations.

    [edit] our posts seem to have crossed ... what's important is your method.
    (I got 9/52 ≈ 0.17308 different rounding)
     
    Last edited: Oct 7, 2012
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