Using Bernoulli’s Trial P(Success=x) \(=n_{C_x}.p^x.q^{(n-x)}\)
x = 0, 1, 2, ………n and q = (1 - p), n = 8
The probability of success, i.e. the bulb is defective \(=p=\frac{6}{100}=\frac{6}{100}\) \(q=1-\frac{6}{100}=\frac{94}{100}\)
probability of that there is not more than one defective piece =
P(0 defective items) + P(1 defective item) =