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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]

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07 Aug 2012, 07:17
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65% (hard)

Question Stats:

50% (03:37) correct 50% (01:32) wrong based on 36 sessions

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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, 07:45, edited 1 time in total.
Edited the question.

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Re: A & B decide to play a game based on probabilities [#permalink]

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07 Aug 2012, 07:44
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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 $$\frac{\frac{5}{36}}{1-\frac{25}{36}}=\frac{5}{11}$$.

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]

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08 Aug 2012, 09:58
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]

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08 Aug 2012, 10:06
Thanks for the solution

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Re: A and B decide to play a game based on probabilities. They [#permalink]

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03 Oct 2017, 13:41
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Re: A and B decide to play a game based on probabilities. They   [#permalink] 03 Oct 2017, 13:41
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