Question

An inventory control employee selected a random sample of 10 items stored in a warehouse containing hundreds of items. If 98% of all the items in the warehouse have an actual count that matches the recorded count: 

(a) What is the probability that all 10 of the items selected have actual counts which match the recorded counts? 

(b) What is the probability that 8 or more of the items selected have actual counts which match the recorded counts?

Answer 1

Binomial distribution: P(X) = nCx p^x q^n-x

Combination formula: nCr = n!/(r! x (n-r)!)

Sample size, n = 10

P(actual count matches recorded count), p = 0.98

q = 1 - p = 0.02

a) P(all 10 items have actual count which match with the recorded count) = P(10)

= 0.98¹⁰

= 0.8171

b) P(8 or more items have actual count which match with the recorded count) = P(8) + P(9) + P(10)

= 10C8 x 0.98⁸ x 0.02² + 10C9 x 0.98⁹ x 0.02 + 0.98¹⁰

= 45 x 0.98⁸ x 0.02² + 10 x 0.98⁹ x 0.02 + 0.98¹⁰

= 0.0153 + 0.1667 + 0.8171

= 0.9991