An electronics store receives a shipment of 28 e-readers. Unbeknownst to the store, 5 of the e-readers are defective. If the store sells 13 of these e-readers the first day, what is the probability that at least 1 of the 13 buyers will get a defective e-reader? (Round your answer to three decimal places.)

To solve this problem, we first need to calculate the probability of not getting a defective e-reader, and then subtract that from 1 to find the probability of getting at least one defective e-reader.

Step 1: Calculate the total number of ways to choose 13 e-readers from 28. This is a combination problem, so we use the combination formula: C(n, k) = n! / [k!(n-k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial.

So, C(28, 13) = 28! / [13!(28-13)!] = 28! / [13!15!] = 10,348,520 ways.

Step 2: Calculate the number of ways to choose 13 e-readers that are not defective. There are 23 (28-5) non-defective e-readers, so we use the combination formula again:

C(23, 13) = 23! / [13!(23-13)!] = 23! / [13!10!] = 1,144,066 ways.

Step 3: Calculate the probability of not getting a defective e-reader. This is the number of ways to choose 13 non-defective e-readers divided by the total number of ways to choose 13 e-readers.

P(no defective) = 1,144,066 / 10,348,520 = 0.1106.

Step 4: Calculate the probability of getting at least one defective e-reader. This is 1 minus the probability of not getting a defective e-reader.

P(at least one defective) = 1 - P(no defective) = 1 - 0.1106 = 0.8894.

So, the probability that at least 1 of the 13 buyers will get a defective e-reader is approximately 0.889, or 88.9%, when rounded to three decimal places.