Banner Ad 1

Collapse

Announcement

Collapse
No announcement yet.

conditional probability problem

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • conditional probability problem

    Q :A package of 6 fuses are tested where the probability an individual fuse is
    defective is 0.05.What is the probability that more than one fuse will be defective, given at least one is defective?

    Ans :
    P[more than one fuse will be defective/atleast one is defective]=P[more than one fuse will be defective & atleast one fuse is defective]/P[atleast one fuse is defective]



    P[atleast one fuse is defective]= 1-P[X<1]
    =1-P[X=0]
    =1-(C(6,0)*(0.95)^6)
    =0.2649 is denomenator.

    Please help me calculating the numerator ........
    Last edited by 1P2P3P; May 17 2010, 06:41 PM.

  • #2
    Originally posted by 1P2P3P View Post
    Q :A package of 6 fuses are tested where the probability an individual fuse is
    defective is 0.05.What is the probability that more than one fuse will be defective, given at least one is defective?

    Ans :
    P[more than one fuse will be defective/atleast one is defective]=
    P[more than one fuse will be defective & atleast one fuse is defective]/
    P[atleast one fuse is defective]


    P[atleast one fuse is defective]= 1-P[X<1]
    =1-P[X=0]
    =1-(C(6,0)*(0.95)^6)
    =0.2649

    Please help me calculating the numerator ........
    Are you not understanding why it reduces to the last line?

    BTW you mean | where you wrote /

    Comment


    • #3
      I amnot able to solve the highlighted part which is stated below :
      P[more than one fuse will be defective & atleast one fuse is defective]=???
      Last edited by 1P2P3P; May 17 2010, 06:43 PM.

      Comment


      • #4
        Originally posted by 1P2P3P View Post
        I amnot able to solve the highlighted part which is stated below :
        P[more than one fuse will be defective & atleast one fuse is defective]=???
        Since there's an "AND" separating the 2 statements, both need to be satisfied. If you let X ~ # of def. then you are asking for P(X > 1 AND X >= 1) = P(X > 1)

        Comment


        • #5
          thanks a lot !!!!

          Comment

          Working...
          X