GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 19 Jan 2019, 13:19

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
  • FREE Quant Workshop by e-GMAT!

     January 20, 2019

     January 20, 2019

     07:00 AM PST

     07:00 AM PST

    Get personalized insights on how to achieve your Target Quant Score.
  • Free GMAT Strategy Webinar

     January 19, 2019

     January 19, 2019

     07:00 AM PST

     09:00 AM PST

    Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

Three children, John, Paul, and Ringo, are playing a game. Each child

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
SVP
SVP
User avatar
V
Status: Preparing GMAT
Joined: 02 Nov 2016
Posts: 2034
Location: Pakistan
GPA: 3.39
Premium Member CAT Tests
Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

New post 18 May 2017, 12:05
5
8
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

25% (01:22) correct 75% (00:53) wrong based on 164 sessions

HideShow timer Statistics

Three children, John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number different from those of the other two children, he is declared the winner. If all of the children choose the same number, the process repeats until one child is declared the winner. If Ringo always chooses 2 and the other children select numbers randomly, what is the probability that Ringo is declared the winner?

A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 2/3

_________________

Final days of the GMAT Exam? => All GMAT Flashcards.
This Post Helps = Press +1 Kudos
Best of Luck on the GMAT!!

Manager
Manager
avatar
B
Joined: 03 Oct 2013
Posts: 84
Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

New post 18 May 2017, 12:10
Is the answer 1/2?

I'm thinking that Ringo selects 2. So the other two have to select 1.

so the probability that the first one can selects 1 is 1/2. the second has to select 1. so answer 1/2?
_________________

P.S. Don't forget to give Kudos on the left if you like the solution

CEO
CEO
User avatar
D
Joined: 11 Sep 2015
Posts: 3341
Location: Canada
Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

New post 19 May 2017, 10:05
3
Top Contributor
1
SajjadAhmad wrote:
Three children, John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number different from those of the other two children, he is declared the winner. If all of the children choose the same number, the process repeats until one child is declared the winner. If Ringo always chooses 2 and the other children select numbers randomly, what is the probability that Ringo is declared the winner?

A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 2/3


This is one of my all-time favorite questions!!

The main concept here is that all 3 children are equally likely to win this game (unless one of them possesses supernatural powers that allow him to know what numbers the other two boys will choose :-))

Also note that, if everything is random, the probability of winning by choosing the number 2 is the same as the probability of winning by choosing the number 1.

So, regardless of what number Ringo chooses, his probability of winning is exactly the same as each of the other boys winning.

Since all 3 boys have the same probability of winning, P(Ringo wins) = 1/3

Likewise, P(John wins) = 1/3 and P(Paul wins) = 1/3

Answer:

Cheers,
Brent
_________________

Test confidently with gmatprepnow.com
Image

Current Student
avatar
B
Joined: 09 Oct 2016
Posts: 89
Location: United States
GMAT 1: 740 Q49 V42
GPA: 3.49
Reviews Badge
Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

New post 19 May 2017, 10:13
2
This is one of my all-time favorite questions!!

The main concept here is that all 3 children are equally likely to win this game (unless one of them possesses supernatural powers that allow him to know what numbers the other two boys will choose :-))

Also note that, if everything is random, the probability of winning by choosing the number 2 is the same as the probability of winning by choosing the number 1.

So, regardless of what number Ringo chooses, his probability of winning is exactly the same as each of the other boys winning.

Since all 3 boys have the same probability of winning, P(Ringo wins) = 1/3

Likewise, P(John wins) = 1/3 and P(Paul wins) = 1/3

Answer:

Cheers,
Brent[/quote]

I'm having trouble with this one.

I wrote down the other two outcomes that are possible:

1,2
2,1
2,2
1,1

He only wins if they get 1,1. So how is it not 1/4?

Posted from my mobile device
CEO
CEO
User avatar
D
Joined: 11 Sep 2015
Posts: 3341
Location: Canada
Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

New post 19 May 2017, 11:45
2
Top Contributor
gmathopeful19 wrote:
I'm having trouble with this one.

I wrote down the other two outcomes that are possible:

1,2
2,1
2,2
1,1

He only wins if they get 1,1. So how is it not 1/4?

Posted from my mobile device


Great question!
You are answering a slightly different question.
The question asks us to find P(Ringo is declared the winner)
Your solution is the correct answer to P(Ringo wins the FIRST round of the game).

Keep in mind that it's possible that nobody wins in the first round, in which case they need to play again, until someone is declared the winner.

In fact, P(Ringo wins the FIRST round of the game) = 1/4
P(John wins the FIRST round of the game) = 1/4
P(Paul wins the FIRST round of the game) = 1/4
P(NOBODY wins the FIRST round of the game) = 1/4

Cheers,
Brent
_________________

Test confidently with gmatprepnow.com
Image

Senior Manager
Senior Manager
User avatar
G
Joined: 03 Apr 2013
Posts: 276
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
GMAT ToolKit User
Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

New post 06 Jul 2017, 00:45
1
GMATPrepNow wrote:
gmathopeful19 wrote:
I'm having trouble with this one.

I wrote down the other two outcomes that are possible:

1,2
2,1
2,2
1,1

He only wins if they get 1,1. So how is it not 1/4?

Posted from my mobile device


Great question!
You are answering a slightly different question.
The question asks us to find P(Ringo is declared the winner)
Your solution is the correct answer to P(Ringo wins the FIRST round of the game).

Keep in mind that it's possible that nobody wins in the first round, in which case they need to play again, until someone is declared the winner.

In fact, P(Ringo wins the FIRST round of the game) = 1/4
P(John wins the FIRST round of the game) = 1/4
P(Paul wins the FIRST round of the game) = 1/4
P(NOBODY wins the FIRST round of the game) = 1/4

Cheers,
Brent

This is a question that I never get right. I actually got confused with two possible ways of finding the answers. The first one of course led to a 1/4, by similar reasoning used by others. The other is that Ringo does win any round of the game.
Let me elucidate.

\(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}(\frac{1}{4}+\frac{1}{4}+.....)\)

I can solve this one algebraically but this also wouldn't result in the answer.

Now please also help me with this - How do you define the event?
Is it

Ringo wins the first round?
or
Ringo wins any round?
or
In how many ways can Ringo win the game?
or something else?

VeritasPrepKarishma Bunuel, I request other experts to also chip in.

Many thanks GMATPrepNow Brent
_________________

Spread some love..Like = +1 Kudos :)

Senior Manager
Senior Manager
User avatar
B
Joined: 28 Jun 2015
Posts: 292
Concentration: Finance
GPA: 3.5
Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

New post 06 Jul 2017, 02:15
1
Probability of declaring Ringo a winner = No. of ways John and Paul chooses 1/No. of ways John and Paul chooses a different number

No. of ways John and Paul chooses 1 = 1.
No. of ways John and Paul chooses a different number = [J-1,P2], [P-1,J-2], [J-1,P-1] = 3.

Probability = 1/3. Ans - C.
_________________

I used to think the brain was the most important organ. Then I thought, look what’s telling me that.

GMATH Teacher
User avatar
G
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 619
Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

New post 17 Jan 2019, 04:39
1
SajjadAhmad wrote:
Three children, John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number different from those of the other two children, he is declared the winner. If all of the children choose the same number, the process repeats until one child is declared the winner. If Ringo always chooses 2 and the other children select numbers randomly, what is the probability that Ringo is declared the winner?

A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 2/3

\(? = P\left( {{\rm{Ringo}}\,\,{\rm{wins}}} \right)\)

Brent´s solution explores symmetries and it is probably the GMAT-scope way of dealing with this good question.

In the same "vibe", I prefer the following argument:

1. Ringo´s fixed preference is irrelevant once it is known. (I mean, if he always chooses 1 is exactly the same problem, hence the answer cannot be different.)

2. John and Paul have no edges between them (sure!) and each one has no edge facing Ringo (and vice-versa) . Let´s see from (say) Paul´s perception:

> If John chooses like Ringo (John chooses 2), Paul may win (choosing 1) or may let them go to the next round (choosing 2) :: there is an edge of Paul over Ringo.
> If John does not choose like Ringo (John chooses 1), Paul may be beaten by Ringo (Paul choosing 1) or he may be beaten by John (Paul choosing 2) :: there is an edge of Ringo over Paul.

In short: Paul has no edge over Ringo (and vice-versa), and the same reasoning applies to conclude that John also does not have an edge over Ringo (and vice-versa).


Conclusion: all of them have the same probability of winning the game, hence 1/3 is the correct answer.


Now let´s see a much clearer solution, but in the end I will use the sum of a convergent geometric sequence with infinite number of terms... that´s out of GMAT´s scope...

\({\rm{Total}}\,\,\left( {R,J,P} \right) = \left( {2,J,P} \right)\,\,:\,\,\,1 \cdot 2 \cdot 2 = 4\,\,{\rm{equiprobable}}\,\,{\rm{possibilities}}\,\,{\rm{per}}\,\,{\rm{round}}\,\,\,\left( * \right)\)

\({\left( {R,J,P} \right)_N}\,\,\,{\rm{indicates}}\,\,{\rm{choices}}\,\,{\rm{in}}\,\,{\rm{round}}\,\,N\)

\({\rm{Favorable}}\,\,:\,\,\,\left\{ \matrix{
\,{\left( {R,J,P} \right)_1} = \left( {2,1,1} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,{1 \over 4} \hfill \cr
\,{\left( {R,J,P} \right)_1} = \left( {2,2,2} \right)\,\,\,{\rm{and}}\,\,\,\,{\left( {R,J,P} \right)_2} = \left( {2,1,1} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,{1 \over 4} \cdot {1 \over 4} \hfill \cr
\,{\left( {R,J,P} \right)_1} = \left( {2,2,2} \right)\,\,\,{\rm{and}}\,\,\,\,{\left( {R,J,P} \right)_2} = \left( {2,2,2} \right)\,\,\,{\rm{and}}\,\,\,{\left( {R,J,P} \right)_3} = \left( {2,1,1} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,{1 \over 4} \cdot {1 \over 4} \cdot {1 \over 4} \hfill \cr
\, \ldots \hfill \cr} \right.\)

\(? = {1 \over 4} + {\left( {{1 \over 4}} \right)^2} + {\left( {{1 \over 4}} \right)^3} + \ldots = {{{1 \over 4}} \over {1 - {1 \over 4}}} = {1 \over 3}\)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________

Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

EMPOWERgmat Instructor
User avatar
V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13346
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

New post 17 Jan 2019, 18:23
1
Hi All,

We're told that three children - John, Paul, and Ringo, are playing a game. Each child will choose either the number 1 or the number 2. When one child chooses a number DIFFERENT from those of the other two children, he is declared the winner. If all of the children choose the SAME number, the process repeats until one child is declared the winner. We're told that Ringo ALWAYS chooses 2 and the other children select numbers randomly. We're asked for the probability that Ringo is declared the winner. This question can be approached in a number of different ways. Since the number of possible outcomes is so small, you can actually just list them all out to determine the answer to this question:

Ringo ALWAYS chooses 2...
IF>..
John = 1 and Paul = 1, then Ringo WINS
John = 1 and Paul = 2, then Ringo loses
John = 2 and Paul = 1, then Ringo loses
John = 2 and Paul = 2, then the game is REPLAYED.

We're told that the game is played (and replayed) until there is a winner, so there really aren't 4 outcomes - there are only 3 outcomes. Since each winning outcome is equally likely, Ringo will win 1/3 of them.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

GMAT Club Bot
Re: Three children, John, Paul, and Ringo, are playing a game. Each child &nbs [#permalink] 17 Jan 2019, 18:23
Display posts from previous: Sort by

Three children, John, Paul, and Ringo, are playing a game. Each child

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.