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Math Help - Finding the probability of a random person

  1. #1
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    Finding the probability of a random person

    Of a group of patients having injuries, 28% visit both a physical therapist and a chiropractor and 8% visit neither. Say that the probability of visiting a physical therapist exceeds the probability of visiting a chiropractor by 16%. What is the probability of a randomly selected person from this group visiting a physical therapist?

    When I solve the problem, I total up the percentages 28, 8, and 16. Therefore, I say the probability would be 16/52. But when I compare my answer to the back of the book they have 0.68 which makes no sense to me.
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  2. #2
    Super Member Random Variable's Avatar
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    Let T be the event that someone in the group visits a physical therapist.

    Let C be the event that someone in the group visits a chiropractor.

    then P( T \cup C) = 1 - 0.08 = 0.92

    so 0.28 = P(T) + P(C) - 0.92

    EDIT: 0.28 = P(T) + P(C) - 0.16 - 0.92

    P(C) = 0.68
    Last edited by Random Variable; May 31st 2009 at 04:56 AM. Reason: error
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  3. #3
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    Quote Originally Posted by MathRules! View Post
    Of a group of patients having injuries, 28% visit both a physical therapist and a chiropractor and 8% visit neither. Say that the probability of visiting a physical therapist exceeds the probability of visiting a chiropractor by 16%. What is the probability of a randomly selected person from this group visiting a physical therapist?

    When I solve the problem, I total up the percentages 28, 8, and 16. Therefore, I say the probability would be 16/52. But when I compare my answer to the back of the book they have 0.68 which makes no sense to me.
    The percentage that see someone is 100-8=92%. If X% visit the Physio, and Y% visit the Quack then:

    92=X+Y-28

    Also you are told that:

    X-Y=16

    Now solve these simultaneaous equations to find that X=68%, or the probability that someone sees the Physio is 0.68.

    CB
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  4. #4
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    Quote Originally Posted by MathRules! View Post
    Of a group of patients having injuries, 28% visit both a physical therapist and a chiropractor and 8% visit neither. Say that the probability of visiting a physical therapist exceeds the probability of visiting a chiropractor by 16%. What is the probability of a randomly selected person from this group visiting a physical therapist?

    When I solve the problem, I total up the percentages 28, 8, and 16. Therefore, I say the probability would be 16/52. But when I compare my answer to the back of the book they have 0.68 which makes no sense to me.
    There are many ways to do this. Since I'm hopeless with probability formulas I'll draw a Karnaugh table:

    \begin{tabular}{l | c | c | c} & P & P$\, '$ & \\ \hline C & 0.28 & & a \\ \hline C$\, '$ & x & 0.08 & y \\ \hline & a + 0.16 & & 1 \\ \end{tabular}<br />

    So it's clear that x = a - 0.12.

    Therefore y = x + 0.08 = (a - 0.12) + 0.08 = a - 0.04.

    Therefore a + y = 1 => a + a - 0.04 = 1 => a = 0.52.

    Therefore Pr(P) = a + 0.16 = 0.52 + 0.16 = 0.68.


    But as I said, there are many ways of doing it. No doubt someone else will show a different approach.


    Edit: Obviously as I was typing my final words the prophecy was coming true ....
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  5. #5
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    Can somebody show me step by step how they came up with the answer? I just don't see it yet.
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  6. #6
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    Quote Originally Posted by MathRules! View Post
    Can somebody show me step by step how they came up with the answer? I just don't see it yet.
    For goodness sake! You have been given three different approaches. Choose one. Then state what you still don't understand.
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  7. #7
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    Ok, mr fantastic, in your problem can you explain to me how you derived x = a - 0.12.?
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  8. #8
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    Quote Originally Posted by MathRules! View Post
    Ok, mr fantastic, in your problem can you explain to me how you derived x = a - 0.12.?
    0.28 + x = a + 0.16

    => x = a + 0.16 - 0.28 = a + 0.12.

    Note that Pr(C) = a therefore Pr(P) = a + 0.16 (information given in the question).
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