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

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Math Expert
Joined: 02 Sep 2009
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Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The  [#permalink]

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10 Jun 2018, 05:17
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95% (hard)

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31% (01:42) correct 69% (01:57) wrong based on 143 sessions

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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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10 Jun 2018, 07:57
1
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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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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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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16 Jun 2018, 05:28

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

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16 Jun 2018, 06:38
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rahulkashyap wrote:

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 $$\frac{3}{5}$$ 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..
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Re: Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The  [#permalink]

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16 Jun 2018, 08:47
chetan2u wrote:
rahulkashyap wrote:

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..

Posted from my mobile device
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Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The  [#permalink]

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28 Jun 2018, 00:27
chetan2u wrote:
rahulkashyap wrote:

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..

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?
Math Expert
Joined: 02 Aug 2009
Posts: 7684
Re: Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The  [#permalink]

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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 rahulkashyap

INFO given
1) 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  [#permalink]

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

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22 Oct 2018, 21:31
arvindsiv wrote:
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.

You did not consider draw scenario.
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Re: Vishwanathan Anand & Gary Kasparov play a series of 5 chess games. The   [#permalink] 22 Oct 2018, 21:31
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