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Math Help - Conditional probability problem

  1. #1
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    Conditional probability problem

    Another problem I am working on:

    A survey of consumers in a particular community showed that 10% were dissatisfied with plumbing jobs done in their homes. Half the complaints dealt with plumber A. If plumber A does 40% of the plumbing jobs in the
    town, find the following probabilities:

    (a) A consumer will obtain an unsatisfactory plumbing job, given that the plumber was A.
    (b) Consumer will obtain a satisfactory plumbing job, given that the plumber was A.

    Solution

    a)

    I defined 2 events A and B
    A: obtain unsatisfactory plumbing
    B: the plumber is A

    I am not sure if I find the right P(A and B) for the conditional prob.
    Thats what I got:
    P(A)= 40% or 0.4 P(B)=10% or 0.1

    P(A|B)= 0.05/0.4 = 12.5%


    b)
    100-12.5=87.5%

    Am I on the right track? Thanks
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  2. #2
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    Quote Originally Posted by somestudent2 View Post
    Another problem I am working on:

    A survey of consumers in a particular community showed that 10% were dissatisfied with plumbing jobs done in their homes. Half the complaints dealt with plumber A. If plumber A does 40% of the plumbing jobs in the
    town, find the following probabilities:

    (a) A consumer will obtain an unsatisfactory plumbing job, given that the plumber was A.
    (b) Consumer will obtain a satisfactory plumbing job, given that the plumber was A.

    Solution

    a)

    I defined 2 events A and B
    A: obtain unsatisfactory plumbing
    B: the plumber is A

    I am not sure if I find the right P(A and B) for the conditional prob.
    Thats what I got:
    P(A)= 40% or 0.4 P(B)=10% or 0.1

    P(A|B)= 0.05/0.4 = 12.5%


    b)
    100-12.5=87.5%

    Am I on the right track? Thanks
    I always find it easiest to try and show these things visually using, say, a karnaugh table. I'll let A represent plumber A (that makes more sense to me than letting A represent something else and letting B represent plumber A!!)


    I start off with: \, \begin{tabular}{l | c | c | c}  & A & A$\, '$ & \\ \hline Dissatisfied & a & b & a + b \\ \hline Satisfied & c & d & c + d \\  \hline & a + c & b + d & 1 \\ \end{tabular}


    From the given data it's known that:

    a + c = 0.4 and therefore b + d = 0.6.

    a + b = 0.1 and therefore c + d = 0.9

    a = 0.05 (half of the complaints were about A. Half of 0.1 = 0.05).

    a + b = 0.1 therefore 0.5 + b = 0.1 therefore b = 0.5.


    Update the table: \, \begin{tabular}{l | c | c | c}  & A & A$\, '$ & \\ \hline Dissatisfied & 0.05 & 0.05 & 0.1 \\ \hline Satisfied & c & d & 0.9 \\  \hline & 0.4 & 0.6 & 1 \\ \end{tabular}


    It's not hard now to add in the other values:

    0.05 + c = 0.4 therefore c = 0.35.

    0.05 + d = 0.6 therefore d = 0.55.


    Update the table: \, \begin{tabular}{l | c | c | c}  & A & A$\, '$ & \\ \hline Dissatisfied & 0.05 & 0.05 & 0.1 \\ \hline Satisfied & 0.35 & 0.55 & 0.9 \\  \hline & 0.4 & 0.6 & 1 \\ \end{tabular}


    As a check, note that 0.35 + 0.55 adds up to 0.9, which it should.

    Now you can answer the question and any other question asked:

    (a) From the table: Pr(Dissatisfied|A) = 0.05/0.4 = 1/8.

    (b) From the table: Pr(Satisfied|A) = 0.35/0.4 = 7/8.

    Counter-intuitive results, perhaps ......

    Your answers would be correct if the questions were:

    Find the percentage of consumers that will obtain:

    (a) an unsatisfactory plumbing job, given that the plumber was A.
    (b) a satisfactory plumbing job, given that the plumber was A.

    But you were NOT asked to find percentages, you were asked to find probability..... A probability is always a number between 0 and 1 inclusive ........ Capisce?

    Here's another one for you: Find the probability that plumber A was NOT used, given that the consumer was satisfied.
    Last edited by mr fantastic; January 27th 2008 at 01:00 AM.
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  3. #3
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    Thank You very much for a detailed explaination. I did not know about Karnaugh tables before now.

    Here's another one for you: Find the probability that plumber A was NOT used, given that the consumer was satisfied.
    I guess from your Karnaugh's table it would be 0.55/0.6 = 0.92
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  4. #4
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    Quote Originally Posted by somestudent2 View Post
    Thank You very much for a detailed explaination. Mr F says: You're welcome. And thankyou for your thanks - lovely manners.

    I did not know about Karnaugh tables before now.



    I guess from your Karnaugh's table it would be 0.55/0.6 = 0.92
    Not quite.

    Pr(A'|satisfied) = 0.55/0.9 = 11/18.

    The given condition is 'customer satisfied', so you divide by the total 'customer satisfied' probability (the end number of the 'satisfied' row), that is, 0.9.
    Last edited by mr fantastic; January 27th 2008 at 01:22 PM.
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