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Math Help - complement question..

  1. #1
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    complement question..

    "the probability to buy a suite is 0.22
    the probability to buy a shirt is 0.3
    the probability to buy a tie is 0.28
    the probability to buy a suite and shirt is 0.11
    the probability to buy a suite and tie is 0.14
    the probability to buy a tie and shirt is 0.28

    what is the probability that the customer will not by anything?"

    i was told that the probability that the customer will not by anything
    equals the complement of the probability that he buys at least 1 item
    why???

    as i see it to solve it we need to do
    not buy a shirt AND not buy a tie AND not buy suite=
    (1-0.11)*(1-0.14)*(1-0.1)=0.68

    why am i wrong?
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  2. #2
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    P(Not buying anything) = 1 - P(buying at least one thing)

    P(buying at least one thing) => P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) - P(A \cap B \cap C)

    Where P(A): buy a suite P(B): buy a shirt P(C): buy a tie

    and P(ABC) = 0 (since it's not given, I'm assuming the customer can't buy all three for some odd reason).
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  3. #3
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    why cant we do
    not buy a shirt AND not buy a tie AND not buy suite

    those events are not connected we as i see it

    why we cant do it this way?
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  4. #4
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    P(Not buying anything) is the same as "not buy a shirt AND not buy a tie AND not buy suite"

    Sorry, misread your last post, the reason why you can't do that is because in this example P(A \cap B \cap C) = P(A)P(B)P(C) is not true since the 3 events are not independent.
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  5. #5
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    Quote Originally Posted by transgalactic View Post
    why cant we do
    not buy a shirt AND not buy a tie AND not buy suite
    those events are not connected we as i see it
    why we cant do it this way?
    You can do that way as long as you do it correctly.
    But the process is just as long. You still must use inclusion/exclusion.

    You see \mathcal{P}(A^cB^cC^c)\ne \mathcal{P}(A^c) \mathcal{P}(B^c) \mathcal{P}(C^c) unless the events are independent.
    From the given you can see that the events are not independent.
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  6. #6
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    why they are not independant?

    buying a shirt and buying a tie are different things,they are not the same
    you cant buy a tie and partialy buy a shirt with it
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  7. #7
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    P(buying a shirt and tie) = 0.28

    P(buy a shirt)*P(buy a tie) = 0.3*0.28 = 0.084

    If they were independent then P(buy a shirt)*P(buy a tie) = 0.28
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  8. #8
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    AAAAAAAAAAAAA

    i got it now thanks
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  9. #9
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    how to calculate
    <br />
 P(A \cap B \cap C)<br />
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  10. #10
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    I think it's 0 in this case since the probability of buying all three wasnt given (and the events are not independent). Otherwise, you can't really solve the problem.

    But I could be wrong.
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  11. #11
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    ohh i didnt look very good, it is given 0.06
    thanks
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  12. #12
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    in the next part of the question i am asked to find
    what is the chance of buying exactly one item

    i was told to calculate the chance of buying at least 2 items.

    tie=t
    shirt=s
    suite=su
    it is being translated as:
    (t and s) or (t and su) or (s and su)

    why arent they adding the case when we buy all 3
    why its not like this

    (t and s) or (t and su) or (s and su) or (s and t and su)
    ??
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