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[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 ........

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 ........

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