Question

A business has had a problem with accurate customer checks according to a review of 939 customer checks. This review showed that 12.3% of the customer checks had an error. If a sample of 20 customer checks is selected, what is the probability that:

1. 0 customer checks will contain errors?
2. Exactly four customer checks will contain errors?
3. One or two customer checks will contain errors?
4. More than six customer checks will contain errors?

Answer 1

P(error) = p = 0.123 and q = 1-p = 1-0.123 = 0.877 and sample size, n=20

(a) P(X=0) = 20C0 * (0.123)⁰ * (0.877)²⁰ = 0.0724

(b) P(X=4) = 20C4 * (0.123)⁴ * (0.877)¹⁶ = 0.1358

(c) P(X=1 or 2) = P(X=1) + P(X=2) = [20C1 * (0.123)¹ * (0.877)¹⁹] + [20C2 * (0.123)² * (0.877)¹⁸] = 0.2032+0.2707 = 0.4739

(d) P(X>6) = 1 - P(X<=6) = 1 - [ P(X=0) + P(X=1)+P(X=2) + P(X=3)+P(X=4) + P(X=5)+P(X=6)]

= 1-{[20C0 * (0.123)⁰ * (0.877)²⁰] + [20C1 * (0.123)¹ * (0.877)¹⁹]+ [20C2 * (0.123)² * (0.877)¹⁸]+[20C3 * (0.123)³ * (0.877)¹⁷] + [20C4 * (0.123)⁴ * (0.877)¹⁶ ]+[20C5 * (0.123)⁵ * (0.877)¹⁵] + [20C6 * (0.123)⁶ * (0.877)¹⁴]}

= 1 - 0.9923 = 0.0077

=> P(X>6) = 0.0077