Problem Solving

Halimah_O wrote:In a shipment of 20 cars, 3 are found to be defective. If 4 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

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

Hi Halimah_O,

GMAT questions are almost always built around a pattern of some kind (even if you don't recognize that the pattern is there). With tougher questions, it's sometimes helpful to 'play around' with the prompt to see if you can define the pattern (and thus, take advantage of it and get the correct answer).

Here, we're asked for the probability of getting exactly one defective car... but ANY of the 4 cars could be the defective one. Rather than try to deal with every possible way that this could happen, let's focus on just one option to start...

What's the probability that the first car is the only one that's defective...

(Defect)(Not)(Not)(Not) =
(3/20)(17/19)(16/18)(15/17) = (3)(17)(16)(15)/(20)(19)(18)(17)

While this fraction looks complex, we can simplify it...

(3)(17)(16)(15)/(20)(19)(18)(17) =

(3)(16)(15)/(20)(19)(18) =

(3)(8)(3)/(4)(19)(9) =

(8)/(4)(19) =

(2)/(19)

Now, notice that this is a fraction with a denominator that's 19. There are clearly other ways to get exactly one defective car - and they're likely ALSO going to be fractions with a denominator of 19. Looking at the answer choices, there's really only one answer that makes sense....

Final Answer: C

GMAT assassins aren't born, they're made,
Rich