Answer to Question #311352 in Statistics and Probability for kia

Question

There are 7 defective light bulbs in a bin containing 80 light bulbs. In how many ways can you select at least 6 defective light bulbs when you choose 8 light bulbs at random?

Answer 1

First find the number of ways to choose exactly 6 defective bulbs:

There are $\left( \begin{array}{c} 7\\ 6\\\end{array} \right) =7$ ways to choose defective bulbs and $\left( \begin{array}{c} 80-7\\ 8-6\\\end{array} \right) =\frac{73\cdot 72}{2}=2628$ ways to choose none-defective bulbs. The total is $7\cdot 2628=18396$ ways

Next find the number of ways to choose exactly 7 defective bulbs:

There should be chosen 1 none-defective bulb from 80-7=73, hence there are 73 ways.

The total number of ways to choose at least 6 defective bulbs is 18396+73=18469.