conditional probability problem

[1P2P3P]

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 ........

[NoMoreExams]

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 /

[1P2P3P]

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]=???

[NoMoreExams]

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)

[1P2P3P]

thanks a lot !!!!