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.

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
by Aman verma » Mon Feb 22, 2010 6:38 am
Q: A and B pick up a card at random from a well shuffled pack of cards , one after the other , replacing it every time till one of them gets a heart . If A begins the game, then the probability that the game ends with B is

a) 3/7

b)4/7

c)3/4

d)1/4

e)2/7

Ans[spoiler]a) 3/7[/spoiler]
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1132
Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members
Location: India
Thanked: 64 times
GMAT Score:760

by harsh.champ » Mon Feb 22, 2010 6:46 am
Aman verma wrote:Q: A and B pick up a card at random from a well shuffled pack of cards , one after the other , replacing it every time till one of them gets a heart . If A begins the game, then the probability that the game ends with B is

a) 3/7

b)4/7

c)3/4

d)1/4

e)2/7

Its imp. to get your thought process correct.Here it goes for me:-
At first glance I note that-"Now,here I will get an infinite series."
If in the 2nd chance ,the game ends :- P = 39/52[A chooese any of the other 39 cards) x 13/52(B chooses one of the 13 hearts) = 3/4 x 1/4 =3/16
If in the 4th chance,the game ends:-P = (39/52)^3 x 13/52
If in the 6th chance,the game ends:-P = (39/52)^5 x 13/52

So,we get the series as (39/52 x 13/52)*[ 1 + (39/52)^2 + (39/52)^4 +...]
=(3/4 x 1/4)*[ 1 + (3/4)^2 + (3/4)^4 +...]
Sum of an infinite series = 1st term/(1 - common ratio)
Hence the term in [ ........ ] would be 1/( 1 - 9/16)=16/7
[spoiler]Hene the answer is 3/16 x 16/7 =3/7 Hence A is the answer.[/spoiler]
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker :P
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Tue May 26, 2009 7:24 pm

by gmatwriter » Mon Feb 22, 2010 12:20 pm
Hi,

Sorry can you explain how 9/16 is the common ratio? Shouldn't it be 3/16?

harsh.champ wrote:
Aman verma wrote:Q: A and B pick up a card at random from a well shuffled pack of cards , one after the other , replacing it every time till one of them gets a heart . If A begins the game, then the probability that the game ends with B is

a) 3/7

b)4/7

c)3/4

d)1/4

e)2/7

Its imp. to get your thought process correct.Here it goes for me:-
At first glance I note that-"Now,here I will get an infinite series."
If in the 2nd chance ,the game ends :- P = 39/52[A chooese any of the other 39 cards) x 13/52(B chooses one of the 13 hearts) = 3/4 x 1/4 =3/16
If in the 4th chance,the game ends:-P = (39/52)^3 x 13/52
If in the 6th chance,the game ends:-P = (39/52)^5 x 13/52

So,we get the series as (39/52 x 13/52)*[ 1 + (39/52)^2 + (39/52)^4 +...]
=(3/4 x 1/4)*[ 1 + (3/4)^2 + (3/4)^4 +...]
Sum of an infinite series = 1st term/(1 - common ratio)
Hence the term in [ ........ ] would be 1/( 1 - 9/16)=16/7
[spoiler]Hene the answer is 3/16 x 16/7 =3/7 Hence A is the answer.[/spoiler]
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 60
Joined: Wed Jan 07, 2009 1:56 pm
Thanked: 12 times
GMAT Score:750

by arzanr » Mon Feb 22, 2010 7:43 pm
If in the 2nd chance ,the game ends :- P = 39/52[A chooese any of the other 39 cards) x 13/52(B chooses one of the 13 hearts) = 3/4 x 1/4 =3/16
How do you get (39/52)*(13/52)?

Shouldn't it be (39/52)*(13/51) because when B chooses one of the 13 hearts, A has already picked one and therefore, there are only 51 cards left.

Similarly when analyzing longer sequences I'm not sure if you can raise (39/52) n times since after every turn, the fraction would change as follows:

1. 39/52 38/51
2. 37/50 13/49

and so on...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1