BTGmoderatorRO wrote:In a shipment of 20 cars, 3 are found to be defective. If four cars are selected at random, what is the probability that exactly one of the four will be defective?

(A) 170/1615
(B) 3/20
(C) 8/19
(D) 3/5
(E) 4/5

OA is C
Can an Expert give me the mathematical approach to solve this question.

We are given that in a shipment of 20 cars, 3 are defective. We need to determine the probability that from 4 selected cars, 1 will be defective and 3 will not. Note that we have 17 non-defective cars.

Let's first determine the total number of ways to select 4 cars from a group of 20 cars.

The number of ways to select 4 cars from a group of 20 is:

20C4 = (20 x 19 x 18 x 17)/4! = (20 x 19 x 18 x 17)/(4 x 3 x 2 x 1) = 5 x 19 x 3 x 17

Next, let's determine the number of ways to select 1 defective car from 3 defective cars.

3C1 = 3

Now let's determine the number of ways to select 3 non-defective cars from 17 non-defective cars:

17C3 = (17 x 16 x 15)/3! = (17 x 16 x 15)/(3 x 2 x 1) = 17 x 8 x 5

Thus, the probability of selecting 1 defective car and 3 non-defective cars is:

[(3) x (17 x 8 x 5)]/(5 x 19 x 3 x 17)

Notice that the 5, 3, and 17 all cancel, and we are left with:

8/19

Alternate Solution:

Alternatively, we could consider the cars one at a time. Remember, we are determining the probability of selecting 1 defective car and 3 non-defective cars from 20 total cars.

If we select the defective car first, since there are 3 defective cars and 20 total cars, there is a 3/20 chance that the defective car will be selected. Next, since there are 17 non-defective cars and 19 cars left, there is a 17/19 chance a non-defective car will be selected. Similarly, for the third car chosen, there is a 16/18 chance another non-defective car will be selected. For the final car, there is a 15/17 chance a non-defective car will be selected.

However, there are 4 different ways to select the 1 defective and 3 non-defective cars:

D - N - N - N

N - D - N - N

N - N - D - N

N - N - N - D

Each of these 4 ways has the same probability of occurring. Thus, the total probability is 4 x (3/20 x 17/19 x 16/18 x 15/17) = 8/19.

Answer: C