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Math Help - well driller

  1. #1
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    well driller

    I am posting this problem here since I got to turn it in 2 hours:
    It was originally posted in the probability forum for college.

    the probability is 0.6 that a well driller will find a well of water at a depth less than 100 feet in a certain area. Wells are to be drilled for six new homeowners. Assue that finding a well of water at a depth of less than 100 feet is independent from drilling to drilling and that the probability is 0.6 on every drilling. If X is the number of wells of water found. Find:
    1-P(X>4)
    My solution: P(X>4)=1-B(4;6,0.6)

    2-P(X=4)
    P(X=4)=b(4;6,0.6)

    3-mean
    mean=n*p=3.6

    4-variance
    Var^2=n*p*(1-p)

    5-P(X=mean)
    Here since mean =3.6, I have problem to find P(X=mean)

    my calculator does not give me a approximate value for b(3.6;4,0.6) !
    How should I treat this question?
    Thank you for your time .
    B
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by braddy View Post
    I am posting this problem here since I got to turn it in 2 hours:
    It was originally posted in the probability forum for college.

    the probability is 0.6 that a well driller will find a well of water at a depth less than 100 feet in a certain area. Wells are to be drilled for six new homeowners. Assue that finding a well of water at a depth of less than 100 feet is independent from drilling to drilling and that the probability is 0.6 on every drilling. If X is the number of wells of water found. Find:
    1-P(X>4)
    My solution: P(X>4)=1-B(4;6,0.6)
    P(X>4) = P(6) + P(5) = b(6;6,0.6) + b(5;6,0.6) = 0.6^6 + 6 0.4 0.6^5 ~= 0.23328

    2-P(X=4)
    P(X=4)=b(4;6,0.6)
    yes: b(4; 6, 0.6) = 6!/(4! 2!) ) 0.6^4 0.4^2 ~= 0.31104

    3-mean
    mean=n*p=3.6
    yes.

    4-variance
    Var^2=n*p*(1-p)
    Var = n p (1-p) = 6*0.6*0.4 ~= 1.44

    5-P(X=mean)
    Here since mean =3.6, I have problem to find P(X=mean)

    my calculator does not give me a approximate value for b(3.6;4,0.6) !
    How should I treat this question?
    Since X cannot take non-integer values P(X=3.6) = 0.

    RonL
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  3. #3
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    Thanks captainBlack!
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