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Math Help - Problem about probability

  1. #1
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    Problem about probability

    So you have 3 phone companies: Company A has a 41% share on the market, B has 38% and C has 21%, summing up gives 100%, so they are really the only ones.

    Now they say that 35% of A and B and 30% of C company customers are unsatisfied with the service.

    What is probability of a satisfied customer be connected to company B?

    Can anyone help me solve this, thank you!
    Last edited by Wright; January 21st 2010 at 10:32 AM.
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  2. #2
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    Quote Originally Posted by Wright View Post
    B has 38% ....


    Now they say that 35% of B are unsatisfied with the service.

    What is probability of a satisfied customer be connected to company B?
    I would suggest 35% of 38% will 'stay' connected.

    \frac{35}{100}\times\frac{38}{100}= \dots
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  3. #3
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    Quote Originally Posted by Wright View Post
    So you have 3 phone companies: Company A has a 41% share on the market, B has 38% and C has 21%, summing up gives 100%, so they are really the only ones.

    Now they say that 35% of A and B and 30% of C company customers are unsatisfied with the service.

    What is probability of a satisfied customer be connected to company B?
    I interpreted the question to mean: What proportion of all satisfied customers are connected to company B? Since 65% of A and B customers, and 70% of C customers, are satisfied, the answer should then be \frac{38\times0.65}{41\times0.65 + 38\times0.65 + 21\times0.7}.
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  4. #4
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    Hello, Wright!

    I agree with Opalg . . . this is Conditional Probability.


    You have 3 phone companies: Company A has a 41% share on the market,
    B has 38% and C has 21%, summing up gives 100%, so they are really the only ones.

    Now they say that: 35% of A and B and 30% of C company customers are unsatisfied with the service.

    What is probability of a satisfied customer be connected to company B?

    We have: . \begin{array}{ccc} P(A) &=& 0.41 \\ P(B) &=& 0.38 \\ P(C) &=& 0.21 \end{array}


    Let: . \begin{array}{ccc}S &=& \text{satisfied} \\ U &=& \text{unsatisfied} \end{array}

    Then: . \begin{array}{ccccccc}<br />
P(A_U) &=& 0.35 & \Rightarrow & P(A_S) &=& 0.65 \\<br />
P(B_U) &=& 0.35 & \Rightarrow & P(B_S) &=& 0.65 \\<br />
P(C_U) &=& 0.30 & \Rightarrow & P(C_S) &=& 0.70 \end{array}


    Bayes' Theorem: . P(B\,|\,S) \;=\;\frac{P(B\,\wedge\,S)}{P(S)}\;\;{\color{blue}[1]}


    Numerator: . P(B\,\wedge S) \:=\:(0.38)(0.65) \:=\:0.247\;\;{\color{blue}[2]}


    Denominator:

    . . \begin{array}{ccccccc}P(A \wedge S) &=& (0.41)(0.65) &=& 0.2665 \\<br /> <br />
P(B \wedge S) &=& (0.38)(0.65) &=& 0.2470 \\<br />
P(C \wedge S) &=& (0.21)(0.30) &=& 0.1470 \\ \hline<br />
\text{Total:}& & P(S) &= & 0.6605 & {\color{blue}[3]} \end{array}


    Substitute [2] and [3] into [1]: . P(B\,|\,S) \;=\;\frac{0.247}{0.6605} \;\approx\;37.4\%

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  5. #5
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    Thanks for the kind answers everyone.
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