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A and B decide to play a game based on probabilities. They

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A and B decide to play a game based on probabilities. They [#permalink] New post 07 Aug 2012, 06:17
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Difficulty:

  55% (hard)

Question Stats:

57% (05:30) correct 43% (01:32) wrong based on 23 sessions
A and B decide to play a game based on probabilities. They throw a fair dice alternately, whoever gets a 5 on his throw wins. What is the probability that A wins the game, if it is known that B starts the game?

A. 7/12
B. 5/12
C. 5/11
D. 7/11
E. 1/2
[Reveal] Spoiler: OA

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Last edited by Bunuel on 07 Aug 2012, 06:45, edited 1 time in total.
Edited the question.
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Re: A & B decide to play a game based on probabilities [#permalink] New post 07 Aug 2012, 06:44
Expert's post
SOURH7WK wrote:
A and B decide to play a game based on probabilities. They throw a fair dice alternately, whoever gets a 5 on his throw wins. What is the probability that A wins the game, if it is known that B starts the game?

A. 7/12
B. 5/12
C. 5/11
D. 7/11
E. 1/2


The probability that A wins the game after the first round is \frac{5}{6}*\frac{1}{6} (B must get anything but 5 and A must get 5);

The probability that A wins the game after the second round is (\frac{5}{6}*\frac{5}{6})*(\frac{5}{6}*\frac{1}{6}) (both must get anything but 5 in the first round and A must win in the second round);

The probability that A wins the game after the third round is (\frac{5}{6}*\frac{5}{6})*(\frac{5}{6}*\frac{5}{6})*(\frac{5}{6}*\frac{1}{6});

and so on...

The overall probability that A wins, will be the sum of the above probabilities: \frac{5}{6}*\frac{1}{6}+(\frac{5}{6}*\frac{5}{6})*(\frac{5}{6}*\frac{1}{6})+(\frac{5}{6}*\frac{5}{6})*(\frac{5}{6}*\frac{5}{6})*(\frac{5}{6}*\frac{1}{6})+...

Notice that we have the sum of the terms of geometric progression with the first term equal to \frac{5}{6}*\frac{1}{6}=\frac{5}{36} and the common ratio equal to \frac{5}{6}*\frac{5}{6}=\frac{25}{36}.

Now, the sum of infinite geometric progression with common ratio |r|<1, is sum=\frac{b}{1-r}, where b is the first term.

So, the above sum equals to \frac{\frac{5}{36}}{1-\frac{25}{36}}=\frac{5}{11}.

Answer: C.

P.S. Not a GMAT type of question.
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Re: A and B decide to play a game based on probabilities. They [#permalink] New post 08 Aug 2012, 08:58
wow that's complicated, I had it till the point where the probability of A winning in the 3rd round is:
(5/6 x 5/6) x (5/6 x 5/6) x (5/6 x 1/6), but beyond that you lost me on the Geom Progression.

Bunuel is it important to know GP inside out for the GMAT? I've done Arithmetic Progression of course..
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Re: A and B decide to play a game based on probabilities. They [#permalink] New post 08 Aug 2012, 09:06
Thanks for the solution
Re: A and B decide to play a game based on probabilities. They   [#permalink] 08 Aug 2012, 09:06
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