Let X be the number of defective transistors. Let p be the probability that defective transistor.
P(defective transistor) = 0.04
P(non defective ) = 1- 0.04 = 0.96
A random sample of size 30 is tested for defects and we want to find probability of no defective transistor
P(No defective transistor) = P(X= 0)
Here X follows binomial distribution with parameters n=30 and p=0.04
The Binomial probability function with parameters n and p is
P(X=x) = [tex] (nCx) p^{x} (1-p)^{n-x} [/tex]
P(X=0) = [tex] (nC0) p^{0} (1-p)^{n-0} [/tex]
Using n=30 and p=0.04
P(X=0) = [tex] (30C0) 0.04^{0} (1-0.04)^{30-0} [/tex]
P(X=0) = 1*1*(0.96)^30
P(X=0) = 0.2938
The probability that there are no defective transistors in the sample is 0.2938
