(i) 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 = 5
The probability of success, i.e. the bulb is defective \(=p=\frac{6}{100}=\frac{1}{10}\) \(q=1-\frac{1}{10}=\frac{9}{10}\)
probability of that no bulb is defective piece =
P(0 defective items) =
(ii) 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 = 5
The probability of success, i.e. the bulb is defective \(=p=\frac{6}{60}=\frac{1}{10}\) \(q=1-\frac{1}{10}=\frac{9}{10}\)
probability of that there are exactly 2 defective pieces =
P(2 defective items) =
