Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart 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 563 reviewsTarget Test Prep

Probability Of Shots

Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 21
Joined: Tue Jul 20, 2010 11:54 pmFollowed by: 1 member
by manish.jaipatna » Wed Mar 23, 2011 5:38 pm
Hi All,

Can you please help me in solving below probability question.

Q. A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

A. 1
B. 1/256
C. 81/256
D. 175/256
E. 144/256

Thanks
Manish
Thanks
Manish
"The Purpose Of Life Is A Life Of Purpose"

"Ruk Jaana Nahi Kanhi Tu Haarke"
Top
Join the discussion
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 110
Joined: Sun May 25, 2008 3:45 pm

by ajmoney09 » Wed Mar 23, 2011 5:58 pm
I said A.

1 out of ever 4...shoots 4 he is going to hit it once.
Top
Join the discussion
Junior | Next Rank: 30 Posts
Posts: 21
Joined: Tue Jul 20, 2010 11:54 pmFollowed by: 1 member

by manish.jaipatna » Wed Mar 23, 2011 6:03 pm
ajmoney09,

Can you explain me in more detail as i am very weak in probability questions. Also if you have any site to refer for probability concepts then that would be great.


Thanks
Manish
Thanks
Manish
"The Purpose Of Life Is A Life Of Purpose"

"Ruk Jaana Nahi Kanhi Tu Haarke"
Top
Join the discussion
Junior | Next Rank: 30 Posts
Posts: 29
Joined: Sun Feb 20, 2011 11:03 pm

by coolly01 » Wed Mar 23, 2011 6:04 pm
I pick B
1 probability in 4 shot (average )

=> he has 1/4 probability to hit the target in one shot
Then in 4 successive shots he has : 1/4 x 1/4 x 1/4 x 1/4 probability to hit the target => 1/256
What is the OA? :)
Top
Join the discussion
Master | Next Rank: 500 Posts
Posts: 423
Joined: Fri Jun 11, 2010 7:59 amFollowed by: 2 members
Location: Seattle, WA
Thanked: 86 times

by srcc25anu » Wed Mar 23, 2011 6:40 pm
P(hit) = 1/4
P(no hit) = 3/4
find the prob that he doesnt hit the target in 4 shots thats equal to (3/4)^4
subtract this from 1 to get the desired probability (that he will hit the target)
therefore 1-81/256 = 175/256
IMO D
Top
Join the discussion
Junior | Next Rank: 30 Posts
Posts: 21
Joined: Tue Jul 20, 2010 11:54 pmFollowed by: 1 member

by manish.jaipatna » Wed Mar 23, 2011 6:52 pm
srcc25anu,

You are correct. Thanks a lot. But i am not always getting an approach to answer such question :( . Another such question is

We have to choose four numbers between 1 to 10. What is the probability of getting 2 & 3 from the chosen four numbers?
Thanks
Manish
"The Purpose Of Life Is A Life Of Purpose"

"Ruk Jaana Nahi Kanhi Tu Haarke"
Top
Join the discussion
Master | Next Rank: 500 Posts
Posts: 423
Joined: Fri Jun 11, 2010 7:59 amFollowed by: 2 members
Location: Seattle, WA
Thanked: 86 times

by srcc25anu » Wed Mar 23, 2011 7:14 pm
we find the prob of not selecting a 2&3 in 8c4 / 10c4 = 1/3
subtract this from 1 to get the desired probability. that should be 1-1/3 = 2/3
manish.jaipatna wrote:srcc25anu,

We have to choose four numbers between 1 to 10. What is the probability of getting 2 & 3 from the chosen four numbers?
Top
Join the discussion
User avatar
Master | Next Rank: 500 Posts
Posts: 436
Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members
Thanked: 72 times

by manpsingh87 » Wed Mar 23, 2011 7:37 pm
manish.jaipatna wrote:Hi All,

Can you please help me in solving below probability question.

Q. A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

A. 1
B. 1/256
C. 81/256
D. 175/256
E. 144/256

Thanks
Manish
probability of hitting the target=1/4;
probability of not hitting the target=3/4;

H- Hit the target
M- Miss the target

Required probability = H+MH+MMH+MMMH
=1/4+3/4*(1/4)+(3/4)*(3/4)*(1/4)+(3/4)*(3/4)*(3/4)*(1/4);
=64+48+36+27/256
=175/276

hence D
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
Top
Join the discussion