BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Hard Probability

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 14
Joined: Thu Sep 19, 2013 3:56 pm

Hard Probability

by Halimah_O » Mon Dec 05, 2016 6:30 am
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
Top
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Mon Dec 05, 2016 8:31 am
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
Let D = defective and N = not defective.

P(exactly n times) = P(one way) * total possible ways.

P(one way):
One way to get exactly one D:
DNNN.
P(D on the 1st pick) = 3/20. (Of the 20 cars, 3 are defective.)
P(N on the 2nd pick) = 17/19. (Of the 19 remaining cars, 17 are not defective.)
P(N on the 3rd pick) = 16/18. (Of the 18 remaining cars, 16 are not defective.)
P(N on the last pick) = 15/17. (Of the 17 remaining cars, 15 are not defective.)
Since we want all of these events to happen, we MULTIPLY:
3/20 * 17/19 * 16/18 * 15/17 = 2/19.

Total possible ways:
DNNN is only ONE WAY to get exactly one D.
Now we must account for ALL OF THE WAYS to get exactly one D.
Any arrangement of the letters DNNN represents one way to get exactly one D.
Thus, to account for ALL OF THE WAYS to get exactly one D, the result above must be multiplied by the number of ways to arrange the letters DNNN.
Number of ways to arrange 4 elements = 4!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 3! to account for the three identical N's:
4!/(3!) = 4.

Multiplying the results above, we get:
P(exactly one D) = 2/19 * 4 = 8/19.

The correct answer is C.

More practice:
https://www.beatthegmat.com/select-exact ... 88786.html
https://www.beatthegmat.com/probability- ... 14250.html
https://www.beatthegmat.com/a-single-par ... 28342.html
https://www.beatthegmat.com/at-a-blind-t ... 20058.html
https://www.beatthegmat.com/rain-check-t79099.html
https://www.beatthegmat.com/probability-t227448.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Junior | Next Rank: 30 Posts
Posts: 14
Joined: Thu Sep 19, 2013 3:56 pm

by Halimah_O » Mon Dec 05, 2016 9:29 am
GMATGuruNY wrote:
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
Let D = defective and N = not defective.

P(exactly n times) = P(one way) * total possible ways.

P(one way):
One way to get exactly one D:
DNNN.
P(D on the 1st pick) = 3/20. (Of the 20 cars, 3 are defective.)
P(N on the 2nd pick) = 17/19. (Of the 19 remaining cars, 17 are not defective.)
P(N on the 3rd pick) = 16/18. (Of the 18 remaining cars, 16 are not defective.)
P(N on the last pick) = 15/17. (Of the 17 remaining cars, 15 are not defective.)
Since we want all of these events to happen, we MULTIPLY:
3/20 * 17/19 * 16/18 * 15/17 = 2/19.

Total possible ways:
DNNN is only ONE WAY to get exactly one D.
Now we must account for ALL OF THE WAYS to get exactly one D.
Any arrangement of the letters DNNN represents one way to get exactly one D.
Thus, to account for ALL OF THE WAYS to get exactly one D, the result above must be multiplied by the number of ways to arrange the letters DNNN.
Number of ways to arrange 4 elements = 4!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 3! to account for the three identical N's:
4!/(3!) = 4.

Multiplying the results above, we get:
P(exactly one D) = 2/19 * 4 = 8/19.

The correct answer is C.

More practice:
https://www.beatthegmat.com/select-exact ... 88786.html
https://www.beatthegmat.com/probability- ... 14250.html
https://www.beatthegmat.com/a-single-par ... 28342.html
https://www.beatthegmat.com/at-a-blind-t ... 20058.html
https://www.beatthegmat.com/rain-check-t79099.html
https://www.beatthegmat.com/probability-t227448.html


Thank you sir.
Top
GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pm
Location: East Bay all the way
Thanked: 625 times
Followed by:119 members
GMAT Score:780

by Matt@VeritasPrep » Thu Dec 08, 2016 8:15 pm
Here's an approach that's much shorter, at least to set up:

Probability = (# of target outcomes) / (# of possible outcomes)

We need three good cars and one bad one. There are (17 choose 3) ways to pick three good ones, and (3 choose 1) ways to pick one bad one. So we have (17 choose 3) * (3 choose 1) groups of four.

That takes care of the target outcomes, so let's move to the possible ones. Those would be any way of choosing four cars from our 20, or (20 choose 4).

That gives us (17 choose 3) * (3 choose 1) / (20 choose 4), which (after a little arithmetic) resolves to 8/19.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 235
Joined: Wed Oct 26, 2016 9:21 pm
Thanked: 3 times
Followed by:5 members

by Anaira Mitch » Thu Jan 05, 2017 7:06 pm
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.
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8083
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

by Scott@TargetTestPrep » Mon Jan 09, 2017 9:39 am
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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pm
Location: Palo Alto, CA
Thanked: 2867 times
Followed by:511 members
GMAT Score:800

by [email protected] » Mon Jan 09, 2017 10:59 am
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
Contact Rich at [email protected]
Image
Top
Post new topic Post Reply
• Page 1 of 1