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
