Question

10 items have been selected from a production line to be inspected. Suppose the process is out of control and a defective item just as likely to occur as a non-defective item. Find the probability of the event Observe at least one defective item.

Answer 1

This is a direct application of binomial distribution.

Let X be a number of defective item among 10 items.

P(defective) = P(non-defective) = p = 0.5

Here, X ~ Binomial ( n = 10, p =0.5)

probability mass function of X is,

P(X = x) = ⁿC_x p^x (1-p)^n-x

We want to find, P(X >=1)

P(X >=1)

= 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - [ ¹⁰C₀ * (0.5)⁰ * (1 - 0.5)^10-0 ]

= 1 - [ 1 * 1 * (0.5)¹⁰ ]

= 1 - 0.0009765625

= 0.9990234375

P(X >=1) = 0.9990234375

P(X >=1) = 0.9990 ( rounded to 4 decimal places)

Hence, the probability of the event Observe at least one defective item is 0.9990