MGMAT Challenge Problem 11/20/06: Defective Cars

[brent.r.baumann]

11/20/06
Question
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


The answer is (C) 8/19. I understand the methodology used in the official MGMAT explanation which I will post below. However, I feel that an approach using factorials should work as well. In this case, the denominator of the probability I found to be [20!/(4!)(16!)]. This simplifies to the prime numbers 3*5*17*19.

The numerator I calculated is [17!/(3!*14!)] You can assume for the numerator that you've already chosen the defective vehicle and so I found the number of combinations of the non-defective vehicles.

So, the final probability that I get is [17!/(3!*14!)]/[20!/(4!*16!)]. All simplified out I get to 8/(19*3)= 8/57. The answer is 8/19, so somewhere I'm not finding a 3 to cancel out.

Can anybody help me here? Do I need to multiple the numerator by 3 because there are 3 different defective vehicles to choose from? This would take care of my additional 3 but I don't know why it works. Thanks for help!!

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Official MGMAT explanation
There are four possible ways to pick exactly one defective car when picking four cars: DFFF, FDFF, FFDF, FFFD (D = defective, F = functional).

To find the total probability we must find the probability of each one of these scenarios and add them together (we add because the total probability is the first scenario OR the second OR...).

The probability of the first scenario is the probability of picking a defective car first (3/20) AND then a functional car (17/19) AND then another functional car (16/18) AND then another functional car (15/17).

The probability of this first scenario is the product of these four probabilities:
3/20 x 17/19 x 16/18 x 15/17 = 2/19

The probability of each of the other three scenarios would also be 2/19 since the chance of getting the D first is the same as getting it second, third or fourth.

The total probability of getting exactly one defective car out of four = 2/19 + 2/19 + 2/19 + 2/19 = 8/19.

[georgepa]

Number of ways to choose 4 cars from 20
= 20 Choose 4 = 4845 = 3*5*17*19

Number of ways to choose 3 non defective cars if you have chosen 1 defective car
17 Choose 3 = 680 = 17 * 8 * 5

Number of ways to choose 1 defective car = 3

Therefore: P(Choosing exactly 1 defective car)

= Number of ways of choosing a defective car * Number of ways of choosing 3 non defective cars / Number of ways of choosing 4 cars

= 3 * 17 * 8 * 5 / 3 * 5 * 17 * 19

= 8 / 19

[Ben Ku]

The numerator I calculated is [17!/(3!*14!)] You can assume for the numerator that you've already chosen the defective vehicle and so I found the number of combinations of the non-defective vehicles.

Georgepa hit the nail on the head. For the numerator, you just found different ways to select three NON-defective cars. You need to multiply it by 3 to find the ways to select one defective car. That's where you're missing the factor of 3.

Hope that helps.

[himanshu.hpr]

There are four possible ways to pick exactly one defective car when picking four cars: DFFF, FDFF, FFDF, FFFD (D = defective, F = functional).


I'm having trouble with specifying it in 4 different ways, why is the order important as we are just selecting randomly 4 cars ... so DFFF, FDFF, FFDF, FFFD should accordingly be just 1 case .. ???

[tim]

It has to do with the way the probability is calculated in the explanation - four cars chosen in a very specific order. There are other ways to solve this one that do not depend on order, but the main lesson here is to make sure you know exactly what you are calculating and why..