Question

The internal auditor for your company believes that 10 percent of their invoices contain errors. To check this theory, 20 invoices are randomly selected.
a. What is the probability that of the 20 invoices selected, exactly 5 would contain errors (round to four decimal places)?
b. What is the probability that of the 20 invoices selected, 5 or more would contain errors (round to four decimal places)?

Answer 1

For this problem, the working equation would be the one usedfor repeated trials probability.

Probability = n!/r!(n - r)! * p^r * q(n - r)

The probability of success is p. If we treat the error as success, then p = 0.10 and consequently q= 1 - p = 0.9. The n is the number of trials equal to 20. Then, the r i the number of successes.

a. n=20, p = 0.10, q = 0.9, r = 5
Probability = 20!/5!(20 - 5)! * 0.10^5 * 0.9^(20-5) = 0.0319

b. n=20, p = 0.10, q = 0.9
For this part, you will have to add when r = 5, r=6, r=7 until r=20. That's a tedious process. So, it would be more reasonable to take r = 4, r = 3, r = 2, r = 1 and r = 0. Then, we will just find the difference.

For r=4,
P = 20!/4!(20 - 4)! * 0.10^4 * 0.9^(20-4) = 0.0898
For r=3,
P = 20!/3!(20 - 3)! * 0.10^3 * 0.9^(20-3) = 0.1901
For r=2,
P = 20!/2!(20 - 2)! * 0.10^2 * 0.9^(20-2) = 0.2852
For r=1,
P = 20!/1!(20 - 1)! * 0.10^1 * 0.9^(20-1) = 0.2702
For r=0,
P = 20!/0!(20 - 0)! * 0.10^0 * 0.9^(20-0) = 0.1216

Total P = 1 - (0.0898+0.1901+0.2852+0.2702+0.1216)
Total P = 0.2332