every time i try to solve this with the formula I get the answer undefined

Question

A machine that manufactures automobile parts produces defective parts 12%

 of the time. If 8 parts produced by this machine are randomly selected, what is the probability that at most 2

 of the parts are defective?

Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.

Answer 1

P(d) = 0.12

P(d<3 of 8) = ?

P(d=0) = 8C0(.88) =8!/8!0! = (.88)^8

P(d=1) == 8C1(.88)^7(.12) = 8!/7!1!(.88)^7(.12) = 8(.88)^7(.12)

P(d=2) = 8C2(.88)^6(.12)^2 = 8!/6!/2!(.88)^6(.12)^2= 28(.88)^6(.12)^2

P(d=r) = 8Cr(.88)^(n-r)(.12)^r = 8!/r!(n-r)!(.88)^(n-r)(.12)^r

P(d<3) = Probability of at most 2 defects in 8 = P(d=0)+P(d=1) + P(d=2)

= .88^8 + 8(.88)^7(.12) + 28(.88)^6(.12)^2

=about .3596 + .3923 + .1872

= .about .9391

= about 94% chance of no more than 2 defectives of a sample of 8

= probability of either 0, 1, or 2 defectives