Answer:
[tex]1 - \frac{(20-k)(19-k)(18-k)(17-k)(16-k)}{20*19*18*17*16}[/tex]
Step-by-step explanation:
If k items out of 20 total items are defective, then the number of non-defective item is 20 – k. The probability of choosing 5 items where all of them are non-defective is
- For the 1st slot the chance is (20 – k)/20
- For the 2nd slot the chance is (20-k-1)/(20-1) = (19-k)/19
- For the 3rd slot: (18 – k)/18
- 4th : (17 – k)/17
- 5th: (16 – k)/16
The probability in total would be
[tex]\frac{(20-k)(19-k)(18-k)(17-k)(16-k)}{20*19*18*17*16}[/tex]
So the probability of selecting at least 1 defective item is the inverse of this, which is
[tex]1 - \frac{(20- k)(19-k)(18-k)(17-k)(16-k)}{20*19*18*17*16}[/tex]
