Question

A manufacturing company performs a quality-control analysis on the ceramic tile it produces. Suppose a batch of 21 tiles has 7 defective tiles. If 6 tiles are sampled at random, what is the probability that exactly 1 of the sampled tiles is defective? How many ways can 6 tiles be selected from 21 tiles? ways (Type a whole number.)

Answer 1

Using the hypergeometric probability distribution, the probability that exactly one out of six sampled tiles is defective is approximately 0.2583. There are 54264 ways to select 6 tiles from a batch of 21 tiles.

Step-by-step explanation:

The question presented is a probability question concerning a batch of ceramic tiles with some being defective. To find the probability that exactly one of the sampled tiles is defective, we can use the hypergeometric probability distribution since the sampling is without replacement. The hypergeometric probability is given by:

P(X=k) = [(C(D, k) * C(N-D, n-k)) / C(N, n)]

where:

C(a, b) is the combination of a items taken b at a time.

D is the number of defective tiles in the batch.

N is the total number of tiles in the batch.

n is the number of tiles sampled.

k is the number of defective tiles found in the sample.

Here, D = 7, N = 21, n = 6, and k = 1. Using the combinations, we can calculate the probability:

P(X=1) = [(C(7, 1) * C(14, 5)) / C(21, 6)]
= [(7! / (1! * (7 - 1)!)) * (14! / (5! * (14 - 5)!))] / (21! / (6! * (21 - 6)!))
= [(7 * 14! / (5! * 9!))] / (21! / (6! * 15!))
= (7 * 2002) / 54264
= 14014 / 54264
\approx 0.2583

The probability that exactly one of the sampled tiles is defective is approximately 0.2583 (25.83%).

To answer the second part of the question, the number of ways to select 6 tiles from 21 is given by the combination formula C(21, 6):

C(21, 6) = 21! / (6! * (21 - 6)!)
= 21! / (6! * 15!)
= 54264

Thus, there are 54264 ways to select 6 tiles from 21 tiles.