The internal auditor for your company believes that 10 percent of their invoices contain errors. To check this theory, 20 invoic
es 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)?
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) = <em>0.0319</em>
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) <em>Total P = 0.2332</em>