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Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The
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10 Jun 2018, 05:17
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28% (01:59) correct 72% (01:59) wrong based on 94 sessions
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Re: Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The
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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
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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
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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
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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 12/5 is 3/5



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Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The
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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 12/5 is 3/5 No.. A game of chess can have three results, win by any of the two or a draw... B tells you \(\frac{3}{5}\) to be the probability of K to win.. It is already given V wins probability is \(\frac{2}{5}\) Since \(\frac{2}{5}+\frac{3}{5}=1\).. 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
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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 12/5 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 Posted from my mobile device



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Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The
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28 Jun 2018, 00:27
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 12/5 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 Hi chetan2u, The question asks the probability of Anand winning the series vs. Gary and mentions that his probability of winning each match is \frac{2}{5}. Going by this, the other \frac{3}{5} is a probability when he doesn't win (tie or loss), no? So, statement 1 should be sufficient in itself?



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Re: Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The
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28 Jun 2018, 01:04
Bunuel wrote: 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. Shruti0805 and rahulkashyapINFO given1) 5 games 2) P of Anand winning a game is 2/5 3) person wins if he wins 3 games..... 4) P is being asked for SERIES Statements: (1) The series ends the moment when any of the two wins 3 matches.P of Anand winning is 2/5 If P of draw is 3/5 and Kasparov winning is 0/5.. different P of Anand winning the series If P of draw is 1/5 and Kasparov winning is 2/5.. different P of Anand winning the series so different answers possible depending on P of draw/Kasparov winning insuff (2) The probability that Kasparov wins a game is 3/5.now this tells us that there is NO possiblity of a DRAW as \(\frac{2}{5}+\frac{3}{5}=1\)... we dont have to calculate since we know the info is sufficient sufficient B
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Re: Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The
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25 Jul 2018, 14:10
chetan2u, I approached it this way. Can you please tell me where I went wrong? P(Anand not winning) = 3/5. Stmt1. Tells us Anand can win in either 3, 4 or 5 games. Therefore P(Anand winning series) = P(win in 3)+P(win in 4)+P(win in 5) each of which can be calculated using given information.




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