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

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

I hate Probability questions..plz help

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 512
Joined: Mon Jun 18, 2012 11:31 pmFollowed by: 20 members
Thanked: 42 times
by sana.noor » Sat Feb 02, 2013 10:17 am
In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?
5/21
3/7
4/7
5/7
16/21

E is the right choice, I know that their are 21 different ways to choose 2 people but How to find 16 is really tough for me.
Work hard in Silence, Let Success make the noise.

If you found my Post really helpful, then don't forget to click the Thank/follow me button. :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Sat Feb 02, 2013 10:24 am
sana.noor wrote:In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?
5/21
3/7
4/7
5/7
16/21
First we need to recognize that the given information tells us that the 7 people consist of:
- a sibling trio
- a sibling pair
- and another sibling pair

Using counting techniques:

For this question, it's easier to find the complement.
So P(not siblings) = 1 - P(they are siblings)

P(they are siblings) = [# of ways to select 2 siblings] / [total # of ways to select 2 people]

# of ways to select 2 siblings
Case a) 2 siblings from the sibling trio: from these 3 siblings, we can select 2 siblings in 3C2 ways (3 ways)
Case b) 2 siblings from first sibling pair: we can select 2 siblings in 2C2 ways (1 way)
Case c) 2 siblings from second sibling pair: we can select 2 siblings in 2C2 ways (1 way)

So, total number of ways to select 2 siblings = 3+1+1 = 5

total # of ways to select 2 people
We have 7 people and we want to select 2 of them
We can accomplish this in 7C2 ways (21 ways)


So, P(they are siblings) = 5/21

This means P(not siblings) = 1 - 5/21
= 16/21 = E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Sat Feb 02, 2013 10:33 am
I posted an explanation here:

https://www.beatthegmat.com/siblings-t66114.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
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 17
Joined: Wed Feb 29, 2012 1:58 am

by ziko » Mon Feb 04, 2013 11:48 pm
If there are 3 people out of 7 who have exactly 3 sublings than there should be 3 sublings (whithin that group of 7), and 4 people who have exactly 1 subling there must be two pairs of sublings.
SO, for me it is much easier when i allocate so letters, let say there are AAABBCC. So the quastion asks for the probability of chosing two people so there will not be any sublings.
Option 1)
First step - The probability to take one person from A's is 3/7 multiplied by 4/6 (the probability to chose the second person from other sublings, so for us it does not matter which one to choose as long as he/she is not from A's)
Second step - the probability to take one person from B's 2/7 multiplied by probability of chosing a second person not from B's, 5/6. We need to multiply this result by 2 because we have two such scenarios, namely with B' and C's.
Final step - Add all the probability cases, because we can have either of them, 12/42+20/42=32/42=16/21

I hope that this explanation hepls!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1