Question

A machine produce 15% defective components.In a sample of 5,drawn at random use the binomial distribution to find the probabilities that :
A) there'll be 4 defective items
B) there'll not be more than 3 defective items.
C) All the items will be non defective

Answer 1

Answer on Question #53813 – Math – Statistics and Probability

A machine produces 15% defective components. In a sample of 5, drawn at random use the binomial distribution to find the probabilities that:

A) There'll be 4 defective items

B) There'll not be more than 3 defective items.

C) All the items will be non-defective

Solution

We use the binomial distribution with $p = 0.15, n = 5$.

A) The probability that there'll be 4 defective items is

$$P(k = 4) = \frac{n!}{k! (n - k)!} p^k (1 - p)^{n - k} = \frac{5!}{4! (5 - 4)!} 0.15^4 (1 - 0.15)^{5 - 4} = 0.00215.$$

B) The probability that there'll not be more than 3 defective items is

$$\begin{array}{l} P(k \leq 3) = 1 - P(k > 3) = 1 - \left(P(k = 4) + P(k = 5)\right) = \\ = 1 - \left(\frac{5!}{4! (5 - 4)!} 0.15^4 (1 - 0.15)^{5 - 4} + \frac{5!}{5! (5 - 5)!} 0.15^5 (1 - 0.15)^{5 - 5}\right) = \\ = 1 - (0.00215 + 0.00008) = 0.99777. \end{array}$$

C) The probability that all the items will be non-defective is

$$P(k = 0) = \frac{5!}{0! (5 - 0)!} 0.15^0 (1 - 0.15)^{5 - 0} = 0.44371.$$

Answer: A) 0.00215; B) 0.99777; C) 0.44371.

www.AssignmentExpert.com