Formulas for three-component set problems

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
This topic has expert replies
User avatar
Master | Next Rank: 500 Posts
Posts: 228
Joined: Sun Jun 29, 2008 8:47 pm
Thanked: 15 times
Followed by:1 members

Formulas for three-component set problems

by amitdgr » Mon Aug 18, 2008 1:27 am
u = union and n = intersection

1. For 3 sets A, B, and C: P(AuBuC) : P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

2. To determine the No of persons in exactly one set : P(A) + P(B) + P(C) – 2P(AnB) – 2P(AnC) – 2P(BnC) + 3P(AnBnC)

3. To determine the No of persons in exactly two of the sets : P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC)

4. To determine the No of persons in exactly three of the sets : P(AnBnC)

5. To determine the No of persons in two or more sets : P(AnB) + P(AnC) + P(BnC) – 2P(AnBnC)

6. To determine the No of persons in at least one set : P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)

(SEE THE ATTACHED PICTURE TO UNDERSTAND THE BELOW MORE CLEARLY)

1. For three sets A, B, and C, P(AuBuC): (A+B+C+X+Y+Z+O)
2. Number of people in exactly one set: ( A+B+C)
3. Number of people in exactly two of the sets: (X+Y+Z)
4. Number of people in exactly three of the sets: O
5. Number of people in two or more sets: ( X+Y+Z+O)
6. Number of people only in set A: A
7. P(A): A+X+Y+O
8. P( AnB): X+O

Hope this helps. Please add if i have missed something.
Attachments
SET_THEORY.jpg
VENN DIAGRAM for 3 component set problems.
Please visit my blog Gmat Tips and Strategies
Source: — Quantitative Reasoning |

GMAT/MBA Expert

User avatar
Site Admin
Posts: 6778
Joined: Mon Feb 13, 2006 8:30 am
Location: Los Angeles, CA
Thanked: 1249 times
Followed by:994 members

by beatthegmat » Fri Aug 22, 2008 9:20 am
Thanks so much for contributing this info!
Beat The GMAT | The MBA Social Network
Community Management Team

Research Top GMAT Prep Courses:
https://www.beatthegmat.com/gmat-prep-courses

Research The World's Top MBA Programs:
https://www.beatthegmat.com/mba/school

Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Wed Mar 24, 2010 7:53 am

by iambroke » Wed Mar 24, 2010 7:55 am
Thanks for the info, but I dont believe point 6 makes much sense...


No of persons in at least one set : P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC) = A+B+C+X+Y+Z+2O


So you're saying its > P(AUBUC)???

Please feel free to contradict or add any missing pieces....

User avatar
GMAT Instructor
Posts: 22
Joined: Mon Mar 15, 2010 2:42 pm
Location: Brooklyn, NY
Thanked: 14 times
Followed by:4 members

by stephen@knewton » Sat Mar 27, 2010 10:54 am
I will let the OP address the question regarding point #6 above, which I suspect may be an innocent typo. But let me put in a very enthusiastic plug for the method which follows the all-caps line in that original post referring to the diagram.

I have found, almost without exception, that what has worked for many of my students (sometimes after a bit of arm-twisting) is to assign DIFFERENT variables to the total numbers of people within groups [here expressed as "P(A)" etc] and to each segment of the venn diagram.

For example, let's pretend we are doing a two-set overlap counting students taking Physics and Biology courses. We'll call the total number of students taking physics P, the total taking biology B, and assign additional variables as follows:

w = number taking ONLY physics
y = number taking ONLY bio
x = number taking BOTH (this is the football-shaped overlap of the venn diagram)
z = number taking NEITHER.

So, using those variables, we can express any total that we need:

P = w + x
B = y + x
PuB: x + y + z
PnB: x

And so on ...

Again, nothing WRONG with assigning only one set of variables and subtracting the overlaps, but I've found that the all-addition method tends to reduce unforced errors significantly. Just one instructor's two cents!

And yes, many thanks to amitdgr for the original post!

Cheers, Steve P.
Stephen
GMAT Instructor
Knewton Inc.

User avatar
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Sun Jul 15, 2012 8:06 am

by vivek.meka » Sun Jul 15, 2012 8:08 am
its X+Y+Z+A+B+C+O I THINK