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Math Help - independent (probability theory)

  1. #1
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    Unhappy independent (probability theory)

    a factory produce a product that can have 3 different faults
    Fault: A,B,C
    The fault is independent (probability theory) of each other.

    X: number of faults that one product can have.

    Determine the density function PX(x) if P(A)= 0,20, P(B)=0,05, P(B)=0,10 ?
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  2. #2
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    Quote Originally Posted by sf1903 View Post
    a factory produce a product that can have 3 different faults
    Fault: A,B,C
    The fault is independent (probability theory) of each other.

    X: number of faults that one product can have.

    Determine the density function PX(x) if P(A)= 0,20, P(B)=0,05, P(B)=0,10 ?
    P(X=0)=(1-0.2)(1-0.05)(1-0.10)

    that is there is no fault A, no fault B and no fault C

    p(X=1)=0.2(1-0.05)(1-0.10)+(1-0.2)0.05(1-0.10)+(1-0.2)(1-0.05)0.10

    that is it is the sum of the probabilities that there is a fault A but no B or C, that there is fault B but no A or C and there is fault C but no A or B

    p(X=2)=  ...

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    P(X=0)=(1-0.2)(1-0.05)(1-0.10)

    that is there is no fault A, no fault B and no fault C

    p(X=1)=0.2(1-0.05)(1-0.10)+(1-0.2)0.05(1-0.10)+(1-0.2)(1-0.05)0.10

    that is it is the sum of the probabilities that there is a fault A but no B or C, that there is fault B but no A or C and there is fault C but no A or B

    p(X=2)=  ...

    CB


    P(X=3)=0,2*0,05*0,10= 0,001
    We get P(X=2)=0,032 and P(X=3)=0,001 is it right?
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