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Math Help - Probability of Australian nominates either cancer or heart disease but not both as m

  1. #1
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    Mar 2011
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    Probability of Australian nominates either cancer or heart disease but not both as m

    Hi there,

    I've just started University level statistics and this question has me stumped:

    A survey conducted by Roy Morgan Research asked 1116 people to
    nominate health issues they consider most important. 60% of
    respondents nominated cancer, and 29% mentioned heart disease.
    Assume that these percentages are true for the population of
    Australia and that 25% of all respondents nominated both cancer and
    heart disease as most important health issues.

    What is the probability that a randomly selected Australian
    nominates either cancer or heart disease but not both as most
    important health issues?

    I'm finding the wording of the question difficult, i.e. what are
    they asking us to find, is it the probability of C or H minus the
    probability of C and H?

    My attempt at the question is as follows: (please note that I'm
    using the letter 'n' to represent the symbol 'and' and I'm using the
    letter 'u' to represent the symbol 'or').

    Let P (C) = Cancer
    Let P (H) = Heart Disease

    So am I trying to find, P (C u H) - P (C n H)?

    P (C u H) = P (C) + P (H) - P (C n H) = 0.6 + 0.29 - 0.25 = 0.64

    Therefore is the solution P (C u H) - P (C n H) = 0.64 - 0.25 = 0.39?

    Thanks for any assistance
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  2. #2
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
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    Mauritius
    Posts
    1,260
    Yes, it's indeed asking for:

    P(C \cup H) - P(C \cap H)
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