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Math Help - Probability Problem...

  1. #1
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    Probability Problem...

     \text{suppose we need to interview 150 blind men in a city(door to door survey).}
    \text{the probability that a house is occupied is 0.73.}
    \text{the probability that the house, if occupied, has at least one man is 0.90.}
    \text{ the probability , given a man occupies the home, the man is blind is 0.15.}
    \text{the likelihood that a blind man completes the interview is 0.95.}

    \text{how many houses do we have to interview in order to have a 75\% chance}
    \text{that  we obtain 150 complete interviews from the blind men?}


    here's what I have attempted:

    \mbox{P(one complete interview is obtained from a house with a blind man)}\;=\; 0.73(0.90)(0.15)(0.95)=0.0937

    \text{no of houses we would need to survey to complete 150 interviews} \;=\; \frac{150}{0.0937} = 1600

    now for the last part ,

    \text{With 75\% chance to obtain 150 interviews, we need,}\;1600 \times 0.75 \;=\;1200

    is this the right approach?
    Last edited by chutiya; October 21st 2011 at 05:14 PM.
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  2. #2
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    Re: Probability Problem...

    Quote Originally Posted by chutiya View Post
    \text{the probability that the house, if occupied, has at least one blind man is 0.90.}
    I'm not convinced this should say "blind".
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  3. #3
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    Re: Probability Problem...

    Given 1600 door knocks, you have a 50% chance of obtaining your required interviews. You need MORE thatn 1600 to get a 75% chance.

    N is the number of doors needed.
    Mean = N * 0.0937 = 150
    Variance = N * 0.0937 * (1-0.0937)

    There's enough that you probably can get away with a Normal Approximation.

    What's next?
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  4. #4
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    Re: Probability Problem...

    Quote Originally Posted by TKHunny View Post
    I'm not convinced this should say "blind".
    Sorry. that was a typo.. it should just be \text{P(house, if occupied, has at least one man)=0.90}


    Quote Originally Posted by TKHunny View Post
    Given 1600 door knocks, you have a 50% chance of obtaining your required interviews. You need MORE thatn 1600 to get a 75% chance.

    N is the number of doors needed.
    Mean = N * 0.0937 = 150
    Variance = N * 0.0937 * (1-0.0937)

    There's enough that you probably can get away with a Normal Approximation.

    What's next?
    I do not understand why we have only a 50% chance of obtaining the required interviews with N =1600?

    And why is it that with N being the number of doors needed, you have mentioned that Mean = N * 0.0937 = 150. Doesn't this mean that N =1600???
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  5. #5
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    Re: Probability Problem...

    Obscure hints aren't helping? I should try to be more clear.

    You need a value, N, such that.

    N*0.0937 = Mean > 150

    N*0.0937*(1-0.0937) = Variance

    SD = sqrt(Variance)

    Mean - 0.67449*SD = 150

    Your challenge, should you choose to accept it, is to tell me where I dreamed up that "-0.67449".
    Last edited by TKHunny; October 21st 2011 at 08:57 PM.
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  6. #6
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    Re: Probability Problem...

    0.67449 is the z value for probability 0.75, right?

    So, I used \text{Mean}- 0.67449*SD = 150 to find N.

    Thank You!!
    Last edited by chutiya; October 21st 2011 at 09:47 PM.
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  7. #7
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    Re: Probability Problem...

    Using a Normal Approximation, that is 75% to the right and 25% to the left. Good.
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  8. #8
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    Re: Probability Problem...

    Finally, I understand this normal approximation thing. Thank you so much for your hints!
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