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
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
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
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.

130-point 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.

Probability boys girls

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 236
Joined: Fri Nov 28, 2008 9:16 am
Thanked: 9 times
by El Cucu » Mon Mar 30, 2009 7:39 am
7 girls 3 boys 10 seats, probability of boys seating separately?

I am a bit lost.

It is ok to calculate the probability of seating together divided by the total outcomes ? Best aproach? tksvm
Last edited by El Cucu on Mon Mar 30, 2009 4:50 pm, edited 1 time in total.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 92
Joined: Wed Mar 25, 2009 12:13 pm
Thanked: 10 times

by moutar » Mon Mar 30, 2009 7:53 am
I'm not very good with probability but I would say...

Total number of ways of sitting 10 people = 10!
Number of ways boys would be sitting together:
BBBBBBBGGG
GBBBBBBBGG
GGBBBBBBBG
GGGBBBBBBB
=4

Prob boys seating separately = 1 - 4/10!

which seems a little steep.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 236
Joined: Fri Nov 28, 2008 9:16 am
Thanked: 9 times

by El Cucu » Mon Mar 30, 2009 8:28 am
moutar wrote:I'm not very good with probability but I would say...

Total number of ways of sitting 10 people = 10!
Number of ways boys would be sitting together:
BBBBBBBGGG
GBBBBBBBGG
GGBBBBBBBG
GGGBBBBBBB
=4

Prob boys seating separately = 1 - 4/10!

which seems a little steep.
Answer is 14/15
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 92
Joined: Wed Mar 25, 2009 12:13 pm
Thanked: 10 times

by moutar » Mon Mar 30, 2009 8:34 am
Let me try again.

Total number of ways of sitting 10 people = 10!
Number of ways boys would be sitting together:
BBBBBBBGGG (7! x 3!)
GBBBBBBBGG
GGBBBBBBBG
GGGBBBBBBB
= 7! x 3! x 4

Prob boys seated together = 7! x 3! x 4/10!
= 2 x 3 x 4/8 x 9 x 10
= 1/30

Prob boys sitting apart = 29/30

I hate probability.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 682
Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member
Thanked: 32 times

Re: Probability boys girls

by Vemuri » Mon Mar 30, 2009 10:00 am
How can only 3 girls separate 7 boys? Doesn't sound logical.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1035
Joined: Wed Aug 27, 2008 10:56 pmFollowed by: 1 member
Thanked: 104 times

by scoobydooby » Mon Mar 30, 2009 10:10 am
agree with Vemuri. 7 boys cant sit separately in 10 seats.

guess the question was about "no girls" being together?

consider the case where all girls are together and take them as a unit of 3 girls. so we have to arrange 7 boys and 1 unit (of 3 girls) or 8 items
in 8! ways. the girls themselves can be arranged in 3! ways within the unit.
so possible arrangements with girls being together 8!*3! ways

10 people can be arranged in 10 seats in 10! ways

cases in which no girls are together: 10!-8!*3!

probabilty of no girls being together: 10!-8!*3!/10!
=1-[(8!*3!)/10!]
=1-3/45
=42/45
=14/15
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 26
Joined: Tue Oct 21, 2008 12:10 pm
Thanked: 3 times
GMAT Score:770

by cm47323 » Mon Mar 30, 2009 12:38 pm
scoobydooby wrote: probabilty of no girls being together: 10!-8!*3!/10!
You solved P(girls not all together), different from P(no girls together). This question was poorly worded from the start...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 236
Joined: Fri Nov 28, 2008 9:16 am
Thanked: 9 times

Re: Probability boys girls

by El Cucu » Mon Mar 30, 2009 4:48 pm
El Cucu wrote:7 boys 3 girls 10 seats, probability of boys seating separately?

I am a bit lost.

It is ok to calculate the probability of seating together divided by the total outcomes ? Best aproach? tksvm
Very sorry for the typo. Is 7 girls and 3 boys, probability of the boys seating separately. ( So I edited the original question)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 682
Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member
Thanked: 32 times

Re: Probability boys girls

by Vemuri » Mon Mar 30, 2009 5:48 pm
El Cucu wrote:
El Cucu wrote:7 boys 3 girls 10 seats, probability of boys seating separately?

I am a bit lost.

It is ok to calculate the probability of seating together divided by the total outcomes ? Best aproach? tksvm
Very sorry for the typo. Is 7 girls and 3 boys, probability of the boys seating separately. ( So I edited the original question)
Hi El, it would be helpful if you can provide the answer options as well.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
MBA Student
Posts: 1194
Joined: Sat Aug 16, 2008 9:42 pmFollowed by: 17 members
Location: Paris, France
Thanked: 71 times
GMAT Score:710

by gmat740 » Mon Mar 30, 2009 9:06 pm
I tried solving it but it looks to me that this question is not fit for GMAT considering the time crunch

Would love to see anybody answer this considering the time restriction

Karan
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 236
Joined: Fri Nov 28, 2008 9:16 am
Thanked: 9 times

by El Cucu » Tue Mar 31, 2009 5:19 pm
gmat740 wrote:I tried solving it but it looks to me that this question is not fit for GMAT considering the time crunch

Would love to see anybody answer this considering the time restriction

Karan
I owe an explanation to all who tried to solve the question.

Well, after 30 minutes of thinking, etc...I consider the problem:

[1- (together/total combinations)] So 1- 8/120= 14/15

Tks for your patience.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1