Question

A bin of 50 parts contains 5 that are defected. A sample of 10 parts is selected at random without replacement. How many samples contain at least four defective parts?

Please show work using factorials.

Answer 1

We want to find atleast four defective parts that means 4 defective or 5 defective parts.

There are 5 parts are defective.

First we will find samples contains exactly 4 defective parts.

The number of ways of selecting 4 parts from the 5 defective parts.

⁵C₄ = 5! / 4!*1! = 5

5 different ways we select 4 defective parts from 5 defective parts.

Now, the number of ways of selecting remaining 6 parts from the 45 non-defective parts.

⁴⁵C₆ = 45! /39!*6! = 8145060.

The number of ways to obtain exactly 4 from 5 defective and 6 from 45 non defective.

⁵C₄ * ⁴⁵C₆ = 5*8145060 = 40725300

Now we will find samples contains exactly 5 defective parts.

The number of ways of selecting 5 parts from the 5 defective parts.

⁵C₅ = 5! / 5!*1! = 1

1 way we select 5 defective parts from 5 defective parts.

Now, the number of ways of selecting remaining 5 parts from the 45 non-defective parts.

⁴⁵C₅ = 45! /40!*5! = 1221759

The number of ways to obtain exactly 5 defective from 5 defective parts and 5 from 45 non defective.

⁵C₅ * ⁴⁵C₅ = 1*1221759 = 1221759

Therefore, the number of samples that contain atleast 4 defective parts is

40725300 + 1221759 = 41947059.

41947059 samples contains atleast four defective parts.