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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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New post Updated on: 07 Aug 2012, 07:45
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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

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Originally posted by SOURH7WK on 07 Aug 2012, 07:17.
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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New post 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 \(\frac{\frac{5}{36}}{1-\frac{25}{36}}=\frac{5}{11}\).

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]

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New post 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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New post 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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