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Three children, John, Paul, and Ringo, are playing a game. Each child
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18 May 2017, 12:05
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25% (01:22) correct 75% (00:53) wrong based on 164 sessions
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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
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child
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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?
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child
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19 May 2017, 10:05
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 alltime 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
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child
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19 May 2017, 10:13
This is one of my alltime 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



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Re: Three children, John, Paul, and Ringo, are playing a game. Each child
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19 May 2017, 11:45
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
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child
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06 Jul 2017, 00:45
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
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child
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06 Jul 2017, 02:15
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 = [J1,P2], [P1,J2], [J1,P1] = 3. Probability = 1/3. Ans  C.
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child
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17 Jan 2019, 04:39
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 GMATscope 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 viceversa) . 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 viceversa), and the same reasoning applies to conclude that John also does not have an edge over Ringo (and viceversa). 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.
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child
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17 Jan 2019, 18:23
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
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Re: Three children, John, Paul, and Ringo, are playing a game. Each child &nbs
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