219k views
1 vote
(Combinatorial analysis question)

A batch of 15 phones contains 3 defective phones. A sample of 4 phones out of 15 in the lot is chosen at random. How many of the possible samples contain at least one defective part?

Book says the answer is 870. Can someone give me a step by step answer please?

User Bobi
by
6.6k points

1 Answer

1 vote

Answer:

Explanation:

There are several ways to approach this problem, but one common method is to use the complementary counting principle, which states that the number of favorable outcomes can be obtained by subtracting the number of unfavorable outcomes from the total number of outcomes. In this case, the favorable outcome is that the sample contains at least one defective phone, and the unfavorable outcome is that the sample contains no defective phones.

The number of possible samples of 4 phones out of 15 can be calculated as the number of combinations of 4 items from a set of 15, which can be written as:

C(15, 4) = 15! / (4! * (15 - 4)!) = 1365

The number of samples containing no defective phones can be calculated as the number of combinations of 4 phones out of the remaining 12 non-defective phones, which can be written as:

C(12, 4) = 12! / (4! * (12 - 4)!) = 495

So, the number of samples containing at least one defective phone can be calculated as:

1365 - 495 = 870

Therefore, out of the 1365 possible samples of 4 phones, 870 of them contain at least one defective phone.

User Muc
by
7.7k points