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

 It is currently 19 Oct 2018, 07:05

Fuqua EA Calls Expected Soon:

Join us in the chat | track the decision tracker | See forum posts/summary

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

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

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

Author Message
TAGS:

Hide Tags

SVP
Status: Preparing GMAT
Joined: 02 Nov 2016
Posts: 1729
Location: Pakistan
GPA: 3.39
Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

18 May 2017, 13:05
5
6
00:00

Difficulty:

95% (hard)

Question Stats:

31% (01:52) correct 69% (01:31) wrong based on 125 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

_________________

Official PS Practice Questions
Press +1 Kudos if this post is helpful

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

Show Tags

18 May 2017, 13:10

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
Joined: 12 Sep 2015
Posts: 3012
Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

19 May 2017, 11:05
3
Top Contributor
1
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

Cheers,
Brent
_________________

Brent Hanneson – GMATPrepNow.com

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

Show Tags

19 May 2017, 11:13
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

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
Joined: 12 Sep 2015
Posts: 3012
Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

19 May 2017, 12:45
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
_________________

Brent Hanneson – GMATPrepNow.com

Senior Manager
Joined: 03 Apr 2013
Posts: 280
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

06 Jul 2017, 01:45
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
Joined: 28 Jun 2015
Posts: 293
Concentration: Finance
GPA: 3.5
Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

06 Jul 2017, 03:15
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.

Non-Human User
Joined: 09 Sep 2013
Posts: 8463
Re: Three children, John, Paul, and Ringo, are playing a game. Each child  [#permalink]

Show Tags

18 Sep 2018, 03:22
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Three children, John, Paul, and Ringo, are playing a game. Each child &nbs [#permalink] 18 Sep 2018, 03:22
Display posts from previous: Sort by