Could I have an explanation of how to solve this question?

Question

A manufacturing machine has a 5% defect rate.

If 9 items are chosen at random, what is the probability that at least one will have a defect?

Answer 1

Pr(d>/=1) = 1- Pr(d = 0)= 100% minus percent chance of No defects

Pr(d=0) = 95% for 1 try. for 9 tries Pr(d=0) = .95^9 = about .63 = 63%

Pr(d>/=1) = about 1-.63 = 100% -63% = about 37% chance of at least 1 defect in 9 tries

or use the last term of the binomial formula expansion

(a+b)^9 with a= .05, b= .95, then last term = .95^9 = about 63%

add up the other terms of the binomial formula expansion, they sum to about 37%

1st term = percent chance of all defects

2nd term = percent chance of 8 defects

3rd term = % chance of 7 defects

...

nth term = % chance of 10-n defects

...

9th term = % chance of 1 defect

10th term = % chance of 0 defects

Answer 2

Assume that each item is independent, use binomial formula:

probability of x successes in n tries when probability of success in each try is p"

x n-x

n! p (1-p)

----

x! (n-x)!

here p = 0.05, n = 9.

probability at least one will have a defect is 1 - probability that none will have a defect.

use above formula for compute probability of 0 detects (x = 0), then subtract that probability from 1.