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Math Help - How can i solve this probability question?

  1. #1
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    How can i solve this probability question?

    In a certain town 45% of the people have black hair,30% have brown eyes and 20% have both black hair and brown eyes.if a person is selected at random from the town,

    What is the probability that he has brown eyes and doesn't have black hair?
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  2. #2
    MHF Contributor matheagle's Avatar
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    Let A=black hair, B=brown eyes.

    P(A)=.45, P(B)=.3 and P(AB)=.2.

    You want P(A'B), where the prime denotes the complement.

    I would use  P(A'B)+P(AB)=P(B).

    SO  P(A'B)+.2=.3, GIVING YOU  P(A'B)=.1.

    My question is where is the probability that this person is male?
    Last edited by matheagle; May 29th 2009 at 10:43 PM.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    Let A=black hair, B=brown eyes.

    P(A)=.45, P(B)=.3 and P(AB)=.2.

    You want P(A'B), where the prime denotes the complement.

    I would use  P(A'B)+P(AB)=P(B).

    My question is where is the probability that this person is male?
    probability=0.3/0.45=0.6

    Is my answer true?
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  4. #4
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by change_for_better View Post
    probability=0.3/0.45=0.6

    Is my answer true?
    I don't agree at all, but if you wanted the conditional probabilities that would be {.2\over .45} and {.2\over .3}

    BUT NOT {.3\over .45}.

    And this is not a conditional (to me) since you have 'brown eyes and doesn't have black hair.'

    AND implies intersection and not a given.
    Last edited by matheagle; May 29th 2009 at 11:12 PM.
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  5. #5
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    Quote Originally Posted by matheagle View Post
    I don't agree at all, but if you wanted the conditional probabilities that would be {.2\over .45} and {.2\over .3}

    BUT NOT {.3\over .45}.

    And this is not a conditional (to me) since you have 'brown eyes and doesn't have black hair.

    AND implies intersection and not a given.
    should I use :

    P(A intersection B)=P(A)*P(B/A)??
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  6. #6
    MHF Contributor matheagle's Avatar
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    why?
    20% have both black hair and brown eyes
    means, to me, P(AB)=.2.

    I left out the basic subtraction earlier today.
    I expected you to finish, but I am tired and I just went back and finished the problem for you.
    All I wanted, was for you to subtract .2 from .3.
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  7. #7
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    Quote Originally Posted by matheagle View Post
    why?
    20% have both black hair and brown eyes
    means, to me, P(AB)=.2.

    I left out the basic subtraction earlier today.
    I expected you to finish, but I am tired and I just went back and finished the problem for you.
    All I wanted, was for you to subtract .2 from .3.

    I think the answer will be :


    P(A)-P(A intersection B)=0.45-0.2=0.25

    Is my answer now true?
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  8. #8
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    Quote Originally Posted by matheagle View Post
    why?
    20% have both black hair and brown eyes
    means, to me, P(AB)=.2.

    I left out the basic subtraction earlier today.
    I expected you to finish, but I am tired and I just went back and finished the problem for you.
    All I wanted, was for you to subtract .2 from .3.
    sorry,I know my mistake now ..

    I'm very sorry..
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  9. #9
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by change_for_better View Post
    probability=0.3/0.45=0.6

    Is my answer true?

    and by the way, what is 30/45?
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  10. #10
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    Quote Originally Posted by change_for_better View Post
    In a certain town 45% of the people have black hair,30% have brown eyes and 20% have both black hair and brown eyes.if a person is selected at random from the town,

    What is the probability that he has brown eyes and doesn't have black hair?
    Here's how I would do that problem. Imagine the town has 1000 residents. Then 30% of 1000= 300 people have brown eyes and 20% of 1000= 200 people have both brown eys and black hair. That leaves 300- 200= 100 who have brown eyes but NOT black hair. The probability of selecting such a person, all people in town being equally likely, is 100/1000= .10.

    And, as matheagle pointed out, .3/.45 is NOT .6!
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  11. #11
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    Quote Originally Posted by change_for_better View Post
    In a certain town 45% of the people have black hair,30% have brown eyes and 20% have both black hair and brown eyes.if a person is selected at random from the town,

    What is the probability that he has brown eyes and doesn't have black hair? Mr F says: He .....??
    Well, if we're showing different approaches then here's mine.

    Make a Karnaugh table (since I'm hopeless with formulas):

    \begin{tabular}{l | c | c | c} & Black hair & Not black hair & \\ \hline Brown eyes & 0.2 & & 0.3 \\ \hline Not brown eyes & & & \\ \hline & 0.45 & & 1 \\ \end{tabular}<br />


    Now it's like a simple Sudoko puzzle to fill in the rest of the probabilities:

    \begin{tabular}{l | c | c | c} & Black hair & Not black hair & \\ \hline Brown eyes & 0.2 & 0.1 & 0.3 \\ \hline Not brown eyes & 0.35 & & 0.7\\ \hline & 0.45 & 0.55 & 1 \\ \end{tabular}<br />


    I'm sure you can fill in the last square.

    So the answer to the question "What is the probability that [the person] has brown eyes and doesn't have black hair?" is easily read from this table to be 0.1.

    However, since no information is given about males and females, the actual question "What is the probability that he has brown eyes and doesn't have black hair?" cannot be answered.
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