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 →

Intermediate Probability

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 304
Joined: Wed Jan 27, 2010 8:35 amFollowed by: 3 members
Location: International Space Station
Thanked: 11 times

Intermediate Probability

by Aman verma » Sun Mar 21, 2010 3:33 am
Q: One hundred identical coins each with probability P of showing up Heads are tossed once. If 0<P<1 and the probability of Heads showing on 50 coins is equal to that of Heads showing on 51 coins, then the value of P is :

a)1/21

b)49/101

c)50/101

d)51/101

e)61/97
800. Arjun's-Bird-Eye
Top
Join the discussion
Log increate an account to reply.
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 189
Joined: Thu Apr 03, 2008 2:03 pm
Location: USA
Thanked: 21 times

by rohan_vus » Sun Mar 21, 2010 4:54 am
Probability of Head for each coin = P
Probability of not having Head for each coin = (1-P)

Probability of 50 Heads = (p)^50*(1-P)^50*100!/(50!*50!) ---- (1)
Probability of 51 Heads = (P)^51*(1-P)^49*100!/(51!*49!) ----(2)

Per question st1 and St2 are same , so (p)^50*(1-P)^50*100!/(50!*50!) = (P)^51*(1-P)^49*100!/(51!*49!)
==> (1-P)/50 = P/51
==> 51 - 51P = 50P
[spoiler]so P = 51/101
D is the answer[/spoiler]
Top
Join the discussion
Log increate an account to reply.
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members
Location: Toronto
Thanked: 1710 times
GMAT Score:800

by Stuart@KaplanGMAT » Sun Mar 21, 2010 1:32 pm
Aman verma wrote:Q: One hundred identical coins each with probability P of showing up Heads are tossed once. If 0<P<1 and the probability of Heads showing on 50 coins is equal to that of Heads showing on 51 coins, then the value of P is :

a)1/21

b)49/101

c)50/101

d)51/101

e)61/97
Without doing any math:

If there was a 1/2 chance of getting heads, then the most frequent number of heads should be 50 out of 100.

In this question, the chance of getting 50 heads is the same as the chance of getting 51 heads; therefore, the coin must be slightly weighted toward heads.

So, we want an answer choice a tiny bit higher than 1/2: only (d) is in the ballpark, choose (d)!
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Join the discussion
Log increate an account to reply.
Master | Next Rank: 500 Posts
Posts: 304
Joined: Wed Jan 27, 2010 8:35 amFollowed by: 3 members
Location: International Space Station
Thanked: 11 times

by Aman verma » Mon Mar 22, 2010 2:01 am
OA D. Thanks to all !
800. Arjun's-Bird-Eye
Top
Join the discussion
Log increate an account to reply.
Senior | Next Rank: 100 Posts
Posts: 46
Joined: Sat Feb 27, 2010 2:58 am
Location: GMAT

by GMATSUCKER » Mon Mar 22, 2010 3:04 am
Stuart Kovinsky wrote:
Without doing any math:

In this question, the chance of getting 50 heads is the same as the chance of getting 51 heads; therefore, the coin must be slightly weighted toward heads.

So, we want an answer choice a tiny bit higher than 1/2: only (d) is in the ballpark, choose (d)!
Love that approach !!!!!!!!!!
GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT, GMAT
What's life without GMAT !!!!!!!!
Top
Join the discussion
Log increate an account to reply.
Junior | Next Rank: 30 Posts
Posts: 29
Joined: Thu Mar 11, 2010 3:23 am
Thanked: 1 times

by Haldiram Bhujiawala » Mon Mar 22, 2010 3:49 am
GMATSUCKER wrote:
Stuart Kovinsky wrote:
Without doing any math:

In this question, the chance of getting 50 heads is the same as the chance of getting 51 heads; therefore, the coin must be slightly weighted toward heads.

So, we want an answer choice a tiny bit higher than 1/2: only (d) is in the ballpark, choose (d)!
Love that approach !!!!!!!!!!
Me Too !!!
Top
Join the discussion
Log increate an account to reply.
Senior | Next Rank: 100 Posts
Posts: 56
Joined: Wed Feb 17, 2010 3:26 am

by mj41 » Tue Mar 23, 2010 9:31 am
rohan_vus can you please explain the equation you used
Top
Join the discussion
Log increate an account to reply.
Post new topic Post Reply
• Page 1 of 1