Question

The items produced by a company contain 10% defective items. Show that the probability of getting 2 defective items in a sample of 8 items is

\[\frac{28 \times 9^6}{{10}^8} .\]

 

Answer 1

Let X denote the number of defective items in the items produced by the company.
Then, X follows binomial distribution with n = 8.

p = 10 % = \[\frac{1}{10}\] 
\[q = 1 - p = \frac{9}{10}\]
\[\text{ Hence, the distribution is given by } \]
\[P(X = r) =^{8}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{8 - r} \]
\[\text{ Prob of getting 2 defective items } = P(X = 2) \]
\[ = ^{8}{}{C}_2 \left( \frac{1}{10} \right)^2 \left( \frac{9}{10} \right)^{8 - 2} \]
\[ = \frac{28 \text{ x } 9^6}{{10}^8}\]