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Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The

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Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The [#permalink]

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New post 10 Jun 2018, 05:17
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Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The probability that Anand wins a game is 2/5. The series will be won by the person who wins 3 matches. Find the probability that Anand wins the series.

(1) The series ends the moment when any of the two wins 3 matches.
(2) The probability that Kasparov wins a game is 3/5.

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Re: Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The [#permalink]

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New post 10 Jun 2018, 07:57
Stmt (1) The series ends the moment when any of the two wins 3 matches.
It is repeating the data already provided in stem,Not sufficient

(2) The probability that Kasparov wins a game is 3/5.
Pbt of kasparov not win = Probability of V.Anand win

Pbt of not winning = 1 - (3/5)
=2/3 hence B , is sufficient
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Re: Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The [#permalink]

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New post 10 Jun 2018, 08:00
Stmt (1) The series ends the moment when any of the two wins 3 matches.
It is repeating the data already provided in stem,Not sufficient

(2) The probability that Kasparov wins a game is 3/5.
Pbt of kasparov not win = Probability of V.Anand win

Pbt of V. Anand win = 1 - (3/5)
=2/3 hence B , is sufficient
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Re: Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The [#permalink]

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New post 16 Jun 2018, 04:43
St.1: Basically, conveys that order matters. So, calculations are based on permutation instead of Combination.

This statement is not sufficient as there could be three scenarios: win, loss and a draw. This is clarified by St.2 by removing the possibility of a draw.

Hence, IMO, Ans is C

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Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The [#permalink]

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New post 16 Jun 2018, 05:28
chetan2u can you please help out with this?

is a saying that the game will end as soon as the winner wins, i.e. it can end in 3/4/5 games? in which case we would need to add up the scenarios in which the winner wins in 3,4 or 5 games.

according to the stem, it seems as though they need to play all 5 games, regardless of when the winner is announced, in which case we would have to calculate permutations to win 3 games out of 5, i.e. multiply by 5!/3!.2!

isnt statement b also implied in the stem? as 1-2/5 is 3/5
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Re: Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The [#permalink]

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New post 16 Jun 2018, 06:38
rahulkashyap wrote:
chetan2u can you please help out with this?

is a saying that the game will end as soon as the winner wins, i.e. it can end in 3/4/5 games? in which case we would need to add up the scenarios in which the winner wins in 3,4 or 5 games.

according to the stem, it seems as though they need to play all 5 games, regardless of when the winner is announced, in which case we would have to calculate permutations to win 3 games out of 5, i.e. multiply by 5!/3!.2!

isnt statement b also implied in the stem? as 1-2/5 is 3/5


No..
A game of chess can have three results, win by any of the two or a draw...
B tells you 3/5 to be the probability of K to win..
It is already given V wins probability is 2/5
Since 2/5+3/5=1..
There is no probability of a draw..
And you can answer now
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Re: Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The [#permalink]

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New post 16 Jun 2018, 08:47
chetan2u wrote:
rahulkashyap wrote:
chetan2u can you please help out with this?

is a saying that the game will end as soon as the winner wins, i.e. it can end in 3/4/5 games? in which case we would need to add up the scenarios in which the winner wins in 3,4 or 5 games.

according to the stem, it seems as though they need to play all 5 games, regardless of when the winner is announced, in which case we would have to calculate permutations to win 3 games out of 5, i.e. multiply by 5!/3!.2!

isnt statement b also implied in the stem? as 1-2/5 is 3/5


No..
A game of chess can have three results, win by any of the two or a draw...
B tells you 3/5 to be the probability of K to win..
It is already given V wins probability is 2/5
Since 2/5+3/5=1..
There is no probability of a draw..
And you can answer now


Can you please help with a detailed solution? I'm still unable to understand your explanation

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Re: Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The   [#permalink] 16 Jun 2018, 08:47
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