NEW HERE? LEARN MORE
closeclose
Last visit was: 27 Apr 2024, 07:16 It is currently 27 Apr 2024, 07:16
Decision Tracker
My Rewards
New posts
New comers' posts

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

Consider three straight lines A, B and C in a plane. 4 unique points a

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
P
Kritisood
Director
Director
Joined: 21 Feb 2017
Posts: 521
Own Kudos [?]: 1042 [16]
Given Kudos: 1091
Location: India
GMAT 1: 700 Q47 V39
Send PM
Reviews Badge
Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink] New post   09 Apr 2020, 08:04
16
Bookmarks
00:00
A
B
C
D
E

Difficulty:

65% (hard)

Question Stats:

56% (02:56) correct 44% (02:52) wrong based on 98 sessions
Hide Show timer Statistics
Consider three straight lines A, B and C in a plane. 4 unique points are marked on line A, 3 unique points are marked on line B and 6 unique points are marked on line C. If we randomly select three marked points, find the probability that the three points form a triangle. Note that no other points, except the points on the same line are collinear

A. 25/286
B. 72/286
C. 90/286
D. 189/286
E. 261/286
ShowHide Answer
Official Answer
E
User avatar
L
chetan2u
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11182
Own Kudos [?]: 31984 [4]
Given Kudos: 291
Send PM
Reviews Badge
Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink] New post   09 Apr 2020, 08:15
2
Kudos
1
Bookmarks
Expert Reply
Kritisood wrote:
Consider three straight lines A, B and C in a plane. 4 unique points are marked on line A, 3 unique points are marked on line B and 6 unique points are marked on line C. If we randomly select three marked points, find the probability that the three points form a triangle. Note that no other points, except the points on the same line are collinear

A. 25/286
B. 72/286
C. 90/286
D. 189/286
E. 261/286




Now making a triangle is SAME as choosing 3 points, so total ways of choosing 3 points out of (4+3+6)=13C3=13*12*11/3!=26*11=286

BUt while choosing these points we have also chosen 3 co-linear points ..
Ways to choose 3 co-linear points = 4C3+3C3+6C3=4+1+20=25

So ways of forming triangles = 286-25=261..
Probability of choosing 261 ways out of 286 ways = 261/286

E
User avatar
D
ArunSharma12
Director
Director
Joined: 25 Oct 2015
Posts: 516
Own Kudos [?]: 879 [0]
Given Kudos: 74
Location: India
Schools: IIML-IPMX '22 (A)
GMAT 1: 650 Q48 V31
GMAT 2: 720 Q49 V38 (Online)
GPA: 4
Send PM
Reviews Badge
Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink] New post   09 Apr 2020, 09:03
Kritisood wrote:
Consider three straight lines A, B and C in a plane. 4 unique points are marked on line A, 3 unique points are marked on line B and 6 unique points are marked on line C. If we randomly select three marked points, find the probability that the three points form a triangle. Note that no other points, except the points on the same line are collinear

A. 25/286
B. 72/286
C. 90/286
D. 189/286
E. 261/286


Total number of triangles formed using points on A,B and C = \(13_C_3\) = 286

total number of triangles formed on Line A = \(4_C_3\) = 4
total number of triangles formed on Line B = \(3_C_3\) = 1
total number of triangles formed on Line C = \(6_C_3\) = 20
we'll have to reduce these triangles from total possible triangles, 286 -25= 261
Probability that the three points form a triangle = \(\frac{261}{286}\)
User avatar
G
preetamsaha
Senior Manager
Senior Manager
Joined: 14 Oct 2019
Status:Today a Reader; Tomorrow a Leader.
Posts: 346
Own Kudos [?]: 344 [0]
Given Kudos: 127
Location: India
GPA: 4
WE:Engineering (Energy and Utilities)
Send PM
Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink] New post   09 Apr 2020, 09:20
total ways of choosing 3 points out of (4+3+6) (because to form a triangle, we need to select any three points.)
=13C3=13*12*11/3!=26*11=286
We have to minus some triangles from those points which are collinear.
number of triangles could have been formed, had these numbers been non-collinear.
= 4C3+3C3+6C3=4+1+20=25
Therefore, the total number of triangles is = 286-25=261..
Probability of choosing 261 ways out of 286 ways = 261/286
correct answer is E
User avatar
P
Kritisood
Director
Director
Joined: 21 Feb 2017
Posts: 521
Own Kudos [?]: 1042 [0]
Given Kudos: 1091
Location: India
GMAT 1: 700 Q47 V39
Send PM
Reviews Badge
Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink] New post   05 May 2020, 01:30
chetan2u wrote:
Kritisood wrote:
Consider three straight lines A, B and C in a plane. 4 unique points are marked on line A, 3 unique points are marked on line B and 6 unique points are marked on line C. If we randomly select three marked points, find the probability that the three points form a triangle. Note that no other points, except the points on the same line are collinear

A. 25/286
B. 72/286
C. 90/286
D. 189/286
E. 261/286




Now making a triangle is SAME as choosing 3 points, so total ways of choosing 3 points out of (4+3+6)=13C3=13*12*11/3!=26*11=286

BUt while choosing these points we have also chosen 3 co-linear points ..
Ways to choose 3 co-linear points = 4C3+3C3+6C3=4+1+20=25

So ways of forming triangles = 286-25=261..
Probability of choosing 261 ways out of 286 ways = 261/286

E


Hi chetan2u i understand your approach completely. I did as follows could you tell me where I faltered?

total points possible = 286
to form a triangle I can choose 1 point from the first line in 4c1 ways, one point from the second line in 3c1 ways, and 1 point from the third line in 6c1 way. Total = 4*3*6=72

I understand i am missing many cases, but why is this approach wrong?
avatar
B
sunnytiss
Current Student
Joined: 02 Mar 2020
Posts: 55
Own Kudos [?]: 14 [0]
Given Kudos: 4
Location: India
Schools: IIMA PGPX "21 (II)
GMAT 1: 740 Q50 V40
Send PM
Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink] New post   05 May 2020, 02:38
1 - (4c3+3c3+6c3) / 13c3 = 261/289
User avatar
L
chetan2u
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11182
Own Kudos [?]: 31984 [3]
Given Kudos: 291
Send PM
Reviews Badge
Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink] New post   05 May 2020, 04:57
2
Kudos
Expert Reply
Kritisood wrote:
chetan2u wrote:
Kritisood wrote:
Consider three straight lines A, B and C in a plane. 4 unique points are marked on line A, 3 unique points are marked on line B and 6 unique points are marked on line C. If we randomly select three marked points, find the probability that the three points form a triangle. Note that no other points, except the points on the same line are collinear

A. 25/286
B. 72/286
C. 90/286
D. 189/286
E. 261/286




Now making a triangle is SAME as choosing 3 points, so total ways of choosing 3 points out of (4+3+6)=13C3=13*12*11/3!=26*11=286

BUt while choosing these points we have also chosen 3 co-linear points ..
Ways to choose 3 co-linear points = 4C3+3C3+6C3=4+1+20=25

So ways of forming triangles = 286-25=261..
Probability of choosing 261 ways out of 286 ways = 261/286

E


Hi chetan2u i understand your approach completely. I did as follows could you tell me where I faltered?

total points possible = 286
to form a triangle I can choose 1 point from the first line in 4c1 ways, one point from the second line in 3c1 ways, and 1 point from the third line in 6c1 way. Total = 4*3*6=72

I understand i am missing many cases, but why is this approach wrong?


What you have taken is when you take one point from each side. Now three points in one line cannot form a triangle but two points on a line can surely make a triangle.
So there will be many cases where you take 2 points on one line and 3rd point on the other line.
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 32717
Own Kudos [?]: 822 [0]
Given Kudos: 0
Send PM
Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink] New post   28 Jun 2023, 11:56
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
gmatclubot
Re: Consider three straight lines A, B and C in a plane. 4 unique points a [#permalink]
28 Jun 2023, 11:56
Moderators:
Bunuel
Math Expert
92952 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions