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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Arrows probability

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 347
Joined: Mon Aug 04, 2008 1:42 pm
Thanked: 1 times
by Stockmoose16 » Wed Sep 17, 2008 10:02 am
If the chances of hitting a target is 1/4, in 5 attempts, what are the chances of hitting the target exactly 2 times?



My method:

(1/4)^2 * (3/4)^3



The correct answer:

(1/4)^2 * (3/4)^3 * 5C2


What's with the 5C2?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 986
Joined: Wed Dec 20, 2006 11:07 amFollowed by: 1 member
Location: India
Thanked: 51 times

by gabriel » Wed Sep 17, 2008 12:16 pm
5c2 is used to calculate the number of ways the arrows hits the target. For eg. it can be the 1st and 2nd time or the 3rd and 4th time. It is similar to choosing 2 things out of 5.

Check out Stuarts post in the GMAT math section. He has given an excellent explanation of a similar problem.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Tue Aug 26, 2008 1:09 am
Thanked: 2 times

by chayanika » Wed Sep 17, 2008 12:21 pm
The 5c2 is the number of ways the 2 times hit can be done in the given 5 attempts.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 347
Joined: Mon Aug 04, 2008 1:42 pm
Thanked: 1 times

by Stockmoose16 » Wed Sep 17, 2008 3:22 pm
chayanika wrote:The 5c2 is the number of ways the 2 times hit can be done in the given 5 attempts.
Why wouldn't the number of ways it can be done be 5!, instead of 5C3?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 829
Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members
Location: INDIA
Thanked: 84 times

Re: Arrows probability

by sudhir3127 » Wed Sep 17, 2008 3:44 pm
Stockmoose16 wrote:If the chances of hitting a target is 1/4, in 5 attempts, what are the chances of hitting the target exactly 2 times?



My method:

(1/4)^2 * (3/4)^3



The correct answer:

(1/4)^2 * (3/4)^3 * 5C2




What's with the 5C2?

Binomial Theroem would be the best way to do such problems

( X+ Y)^n = n C k X^n-k y^k

(https://en.wikipedia.org/wiki/Binomial_theorem)

Thus it will be
5C2 (1/4)^2 ( 1-1/4)^5-2

5C2 * (1/4)^2 * (3/4)^3

Hope that helps...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 347
Joined: Mon Aug 04, 2008 1:42 pm
Thanked: 1 times

Re: Arrows probability

by Stockmoose16 » Wed Sep 17, 2008 3:46 pm
sudhir3127 wrote:
Stockmoose16 wrote:If the chances of hitting a target is 1/4, in 5 attempts, what are the chances of hitting the target exactly 2 times?



My method:

(1/4)^2 * (3/4)^3



The correct answer:

(1/4)^2 * (3/4)^3 * 5C2




What's with the 5C2?

Binomial Theroem would be the best way to do such problems

( X+ Y)^n = n C k X^n-k y^k

(https://en.wikipedia.org/wiki/Binomial_theorem)

Thus it will be
5C2 (1/4)^2 ( 1-1/4)^5-2

5C2 * (1/4)^2 * (3/4)^3

Hope that helps...
Is the reason it's 5C2 rather than 5! because, in this example, there are 3 hits and 2 misses, which would be redundant. So, it's really

5!/3!*2! = 5C2 (The 3!*2! takes out the redundancy of 3 hits and 2 misses)?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 33
Joined: Fri Sep 05, 2008 6:40 am
Thanked: 1 times

by pseudononymous » Wed Sep 17, 2008 6:08 pm
Stockmoose16 wrote:
chayanika wrote:The 5c2 is the number of ways the 2 times hit can be done in the given 5 attempts.
Why wouldn't the number of ways it can be done be 5!, instead of 5C3?
The C2 eliminates duplicates since order doesn't matter.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1