A manufacturing company performs a quality-control analysis on the ceramic tile it produces. Suppose a batch of 21 tiles has 6 defective tiles. If 4 tiles are sampled at random, what is the probability that exactly 1 of the sampled tiles is defective? How many ways can 4 tiles be selected from 21 tiles?

Step 1 ：First, we need to determine the total number of ways to select 4 tiles from a batch of 21. This can be calculated using the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\), where \(n\) is the total number of items, \(k\) is the number of items to choose, and \(!\) denotes factorial. So, the total number of ways to select 4 tiles from 21 is \(C(21, 4)\).

Step 2 ：Calculate \(C(21, 4) = \frac{21!}{4!(21-4)!} = 5985\).

Step 3 ：Next, we need to determine the number of ways to select exactly 1 defective tile from the 6 defective tiles and 3 non-defective tiles from the 15 non-defective tiles. This can be calculated as \(C(6, 1) * C(15, 3)\).

Step 4 ：Calculate \(C(6, 1) = \frac{6!}{1!(6-1)!} = 6\) and \(C(15, 3) = \frac{15!}{3!(15-3)!} = 455\).

Step 5 ：So, the number of ways to select exactly 1 defective tile and 3 non-defective tiles is \(6 * 455 = 2730\).

Step 6 ：The probability that exactly 1 of the sampled tiles is defective is the number of ways to select exactly 1 defective tile and 3 non-defective tiles divided by the total number of ways to select 4 tiles. So, the probability is \(\frac{2730}{5985}\).

Step 7 ：Simplify the fraction to get the final probability. The final probability is \(\boxed{\frac{14}{39}}\).