GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Jun 2019, 08:06 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  In a certain game only one player can win and only one player always

Author Message
TAGS:

Hide Tags

Intern  Joined: 17 Mar 2014
Posts: 28
Location: India
GMAT Date: 07-02-2015
WE: Information Technology (Computer Software)
In a certain game only one player can win and only one player always  [#permalink]

Show Tags

5 00:00

Difficulty:   25% (medium)

Question Stats: 71% (01:23) correct 29% (01:31) wrong based on 253 sessions

HideShow timer Statistics

In a certain game only one player can win and only one player always eventually wins. James, Austin, and Katelyn play this game together 3 times in a row. What is the probability that Katelyn wins at least one of the 3 games.?

(1) The probability that either James or Austin wins the game is 3/4
(2) James and Katelyn have an equal probability of winning the game.

Originally posted by SivaKumarP on 23 Apr 2015, 19:37.
Last edited by Bunuel on 24 Apr 2015, 02:28, edited 1 time in total.
Renamed the topic and edited the question.
Manager  Joined: 28 Jan 2015
Posts: 126
Concentration: General Management, Entrepreneurship
GMAT 1: 670 Q44 V38 Re: In a certain game only one player can win and only one player always  [#permalink]

Show Tags

SivaKumarP wrote:
In a certain game only one player can win and only one player always eventually wins. James, Austin, and Katelyn play this game together 3 times in a row. What is the probability that Katelyn wins at least one of the 3 games.?

1> The probability that either James or Austin wins the game is 3/4
2> James and Katelyn have an equal probability of winning the game.

Probability of anything "at least once" = 1- (probability of it not happening)

Statement 1. Gives sufficient information to equate the above, no need to do the calculation though.

Statement 2. Tells us that James and Katelyn have equal odds but says nothing about the third person, so we cannot derive Katelyn's odds.

Intern  Joined: 17 Mar 2014
Posts: 28
Location: India
GMAT Date: 07-02-2015
WE: Information Technology (Computer Software)
Re: In a certain game only one player can win and only one player always  [#permalink]

Show Tags

Hi Sabineodf,

Yes, you are correct. the answer is A and we need not to calculate. However, since I'm new to probability,after arriving at the answer for the data sufficiency, I just wanted to give a try in calculation to understand my strength in the probability sums. As I doubted, I got a doubt in doing that.

Using the 1st statement, I tried solving the problem

Method 1:

1 - (27/64) gives me 37/64

Method 2:

Katelyn wins the exactly one game = 1/4 * 3/4 * 3/4 =9/64
Katelyn wins exactly twice = 1/4 * 1/4 * 3/4 = 3/64
Katelyn wins exactly 3 thrice = 1/4 * 1/4 * 1/4 =1/64

If we sum up, we get 13/64

can you please tell me where I went wrong? Because I'm getting 2 answers for the same problem Manager  Joined: 28 Jan 2015
Posts: 126
Concentration: General Management, Entrepreneurship
GMAT 1: 670 Q44 V38 Re: In a certain game only one player can win and only one player always  [#permalink]

Show Tags

While I am not sure where you are calculating something wrong, I can show you my work process, I am not fantastic at this but I'm sure someone with a little bit more skill than me will have some input haha.

So, since we know that probability of something happening at least once is 1 - (probability of it not happening) we can calculate as follows

In 1 game the probability that James or Austin wins is 3/4th, so the probability that Katelyn wins in 1 game is 1/4th.

They play three games, so she has 3 chances to win: 1/4 + 1/4 + 1/4 = 3/4

Your calculation of 1 - (27/64) I believe is complicating the problem a little bit, I think you multiplied where you should have added. I did that too initially but then I remembered that her playing the game 3 times means her chances of winning increase, not decrease (as they do when you multiply).

I tried to solve this with logic as my math skills aren't great, but there are probably dozens of ways to get the right answer.. Intern  Joined: 17 Mar 2014
Posts: 28
Location: India
GMAT Date: 07-02-2015
WE: Information Technology (Computer Software)
Re: In a certain game only one player can win and only one player always  [#permalink]

Show Tags

Let me just ellaborate the method 1 a bit more to avoid confusion .

Method 1:

How I arrived at 37/64 is as follows.

In order to use the short cut formula; Probability of getting at least once = 1 -( Probability of not getting anything), we need to calculate that Katelyn losses all the 3 times, which can be calculated by ( 3/4 * 3/4 * 3/4).

1- ( 3/4 * 3/4 * 3/4) = 37/64, which is correct and I've no doubt in that.

While trying the second method without any short cut, there seems to be an issue. I could go with short cut , I shouldnot move on just because I couldnt do it. e-GMAT Representative D
Joined: 04 Jan 2015
Posts: 2888
Re: In a certain game only one player can win and only one player always  [#permalink]

Show Tags

3
1
SivaKumarP wrote:
Hi Sabineodf,

Yes, you are correct. the answer is A and we need not to calculate. However, since I'm new to probability,after arriving at the answer for the data sufficiency, I just wanted to give a try in calculation to understand my strength in the probability sums. As I doubted, I got a doubt in doing that.

Using the 1st statement, I tried solving the problem

Method 1:

1 - (27/64) gives me 37/64

Method 2:

Katelyn wins the exactly one game = 1/4 * 3/4 * 3/4 =9/64
Katelyn wins exactly twice = 1/4 * 1/4 * 3/4 = 3/64
Katelyn wins exactly 3 thrice = 1/4 * 1/4 * 1/4 =1/64

If we sum up, we get 13/64

can you please tell me where I went wrong? Because I'm getting 2 answers for the same problem Hi SivaKumarP,

Let me tell why you have different answers from both the methods:

Method 1:
P(Katelyn winning atleast 1 game) = 1 - P( Katelyn losing all the three games) $$= 1 - (3_{c_3} * \frac{3}{4} * \frac{3}{4} * \frac{3}{4}$$) = $$1 - \frac{27}{64} = \frac{37}{64}$$ which is what you have calculated. Let's see your method 2 now.

Method 2:
P(Katelyn winning atleast 1 game) = P(Katelyn winning only 1 game) +P(Katelyn winning only 2 games) +P(Katelyn winning all 3 games)

= ($$3_{c_1} * \frac{1}{4}* \frac{3}{4} * \frac{3}{4})+ (3_{c_2} * \frac{1}{4} * \frac{1}{4} * \frac{3}{4}) + (3_{c_3} * \frac{1}{4} * \frac{1}{4} * \frac{1}{4})$$

= $$\frac{27}{64} + \frac{9}{64} + \frac{1}{64} = \frac{37}{64}$$ same as your answer in the first method.

You were getting a different answer because you did not choose combination of games which Katelyn would win or lose. So, when Katelyn wins only one game, she may win the 1st game or the 2nd or even the 3rd. Hence, there are 3 ways in which she can win 1 game. Same working would be applicable when Katelyn wins 2 games and all games.

Here method 1 should be the preferred way as it involves lesser no. of scenarios to calculate.

Hope its clear now!

Regards
Harsh
_________________
e-GMAT Representative D
Joined: 04 Jan 2015
Posts: 2888
Re: In a certain game only one player can win  [#permalink]

Show Tags

1
sabineodf wrote:
While I am not sure where you are calculating something wrong, I can show you my work process, I am not fantastic at this but I'm sure someone with a little bit more skill than me will have some input haha.

So, since we know that probability of something happening at least once is 1 - (probability of it not happening) we can calculate as follows

In 1 game the probability that James or Austin wins is 3/4th, so the probability that Katelyn wins in 1 game is 1/4th.

They play three games, so she has 3 chances to win: 1/4 + 1/4 + 1/4 = 3/4

Your calculation of 1 - (27/64) I believe is complicating the problem a little bit, I think you multiplied where you should have added. I did that too initially but then I remembered that her playing the game 3 times means her chances of winning increase, not decrease (as they do when you multiply).

I tried to solve this with logic as my math skills aren't great, but there are probably dozens of ways to get the right answer.. Hi sabineodf,

In case of an AND event, we would multiply the probabilities and not add them. The probabilities are added in case its an OR event. To illustrate my point, let me consider the example in the question.

AND Event
For the event where Katelyn wins all 3 games, P(Katelyn winning all 3 games) = P( Katelyn winning the 1st game AND Katelyn winning the 2nd game AND Katelyn winning the 3rd game) $$= \frac{1}{4} * \frac{1}{4} * \frac{1}{4} = \frac{1}{64}$$

In case I extend your logic of adding the probabilities, for 5 games the probability of Katelyn winning would become $$\frac{5}{4}$$, which is not possible.

OR Event
Example of an OR event would be P(Katelyn winning atleast 1 game) = P(Katelyn winning only 1 game) OR P(Katelyn winning only 2 games) OR P(Katelyn winning all 3 games). In this case we would add the probabilities as I have done in the explanation above.

Hope its clear now!

Regards
Harsh
_________________
Intern  Joined: 17 Mar 2014
Posts: 28
Location: India
GMAT Date: 07-02-2015
WE: Information Technology (Computer Software)
Re: In a certain game only one player can win and only one player always  [#permalink]

Show Tags

EgmatQuantExpert wrote:
SivaKumarP wrote:
Hi Sabineodf,

Yes, you are correct. the answer is A and we need not to calculate. However, since I'm new to probability,after arriving at the answer for the data sufficiency, I just wanted to give a try in calculation to understand my strength in the probability sums. As I doubted, I got a doubt in doing that.

Using the 1st statement, I tried solving the problem

Method 1:

1 - (27/64) gives me 37/64

Method 2:

Katelyn wins the exactly one game = 1/4 * 3/4 * 3/4 =9/64
Katelyn wins exactly twice = 1/4 * 1/4 * 3/4 = 3/64
Katelyn wins exactly 3 thrice = 1/4 * 1/4 * 1/4 =1/64

If we sum up, we get 13/64

can you please tell me where I went wrong? Because I'm getting 2 answers for the same problem Hi SivaKumarP,

Let me tell why you have different answers from both the methods:

Method 1:
P(Katelyn winning atleast 1 game) = 1 - P( Katelyn losing all the three games) $$= 1 - (3_{c_3} * \frac{3}{4} * \frac{3}{4} * \frac{3}{4}$$) = $$1 - \frac{27}{64} = \frac{37}{64}$$ which is what you have calculated. Let's see your method 2 now.

Method 2:
P(Katelyn winning atleast 1 game) = P(Katelyn winning only 1 game) +P(Katelyn winning only 2 games) +P(Katelyn winning all 3 games)

= ($$3_{c_1} * \frac{1}{4}* \frac{3}{4} * \frac{3}{4})+ (3_{c_2} * \frac{1}{4} * \frac{1}{4} * \frac{3}{4}) + (3_{c_3} * \frac{1}{4} * \frac{1}{4} * \frac{1}{4})$$

= $$\frac{27}{64} + \frac{9}{64} + \frac{1}{64} = \frac{37}{64}$$ same as your answer in the first method.

You were getting a different answer because you did not choose combination of games which Katelyn would win or lose. So, when Katelyn wins only one game, she may win the 1st game or the 2nd or even the 3rd. Hence, there are 3 ways in which she can win 1 game. Same working would be applicable when Katelyn wins 2 games and all games.

Here method 1 should be the preferred way as it involves lesser no. of scenarios to calculate.

Hope its clear now!

Regards
Harsh

Hi Harsh,

Awesome explanation. Many kudos.. One morething..3C1 means out of the 3 games she won 1?? Because I usually dont go with this formula as I used to go with some dash _ method. This formula seems to be new to me. can you confirm this one as well
e-GMAT Representative D
Joined: 04 Jan 2015
Posts: 2888
Re: In a certain game only one player can win and only one player always  [#permalink]

Show Tags

1
SivaKumarP wrote:
EgmatQuantExpert wrote:
SivaKumarP wrote:
Hi Sabineodf,

Yes, you are correct. the answer is A and we need not to calculate. However, since I'm new to probability,after arriving at the answer for the data sufficiency, I just wanted to give a try in calculation to understand my strength in the probability sums. As I doubted, I got a doubt in doing that.

Using the 1st statement, I tried solving the problem

Method 1:

1 - (27/64) gives me 37/64

Method 2:

Katelyn wins the exactly one game = 1/4 * 3/4 * 3/4 =9/64
Katelyn wins exactly twice = 1/4 * 1/4 * 3/4 = 3/64
Katelyn wins exactly 3 thrice = 1/4 * 1/4 * 1/4 =1/64

If we sum up, we get 13/64

can you please tell me where I went wrong? Because I'm getting 2 answers for the same problem Hi SivaKumarP,

Let me tell why you have different answers from both the methods:

Method 1:
P(Katelyn winning atleast 1 game) = 1 - P( Katelyn losing all the three games) $$= 1 - (3_{c_3} * \frac{3}{4} * \frac{3}{4} * \frac{3}{4}$$) = $$1 - \frac{27}{64} = \frac{37}{64}$$ which is what you have calculated. Let's see your method 2 now.

Method 2:
P(Katelyn winning atleast 1 game) = P(Katelyn winning only 1 game) +P(Katelyn winning only 2 games) +P(Katelyn winning all 3 games)

= ($$3_{c_1} * \frac{1}{4}* \frac{3}{4} * \frac{3}{4})+ (3_{c_2} * \frac{1}{4} * \frac{1}{4} * \frac{3}{4}) + (3_{c_3} * \frac{1}{4} * \frac{1}{4} * \frac{1}{4})$$

= $$\frac{27}{64} + \frac{9}{64} + \frac{1}{64} = \frac{37}{64}$$ same as your answer in the first method.

You were getting a different answer because you did not choose combination of games which Katelyn would win or lose. So, when Katelyn wins only one game, she may win the 1st game or the 2nd or even the 3rd. Hence, there are 3 ways in which she can win 1 game. Same working would be applicable when Katelyn wins 2 games and all games.

Here method 1 should be the preferred way as it involves lesser no. of scenarios to calculate.

Hope its clear now!

Regards
Harsh

Hi Harsh,

Awesome explanation. Many kudos.. One morething..3C1 means out of the 3 games she won 1?? Because I usually dont go with this formula as I used to go with some dash _ method. This formula seems to be new to me. can you confirm this one as well

Thanks SivaKumarP $$3_{c_1}$$ strictly means selecting 1 item out of 3 items, where the order of selection does not matter. Its written as $$3_{c_1}= \frac{3!}{2!*1!} = 3$$. In this case, it means selecting 1 game out of 3 games which Katelyn would win.

In case the order of selection matters, its written as $$3_{p_1} = \frac{3!}{2!} = 3$$. In this case, both the values are same as 1! = 1

Let me know in case you have trouble understanding this.

Regards
Harsh
_________________
Intern  Joined: 17 Mar 2014
Posts: 28
Location: India
GMAT Date: 07-02-2015
WE: Information Technology (Computer Software)
Re: In a certain game only one player can win and only one player always  [#permalink]

Show Tags

Thanks Harsh. I got it and I will try to become master with that This time instead of saying kudos, I really gave you kudos.. Manager  Joined: 28 Jan 2015
Posts: 126
Concentration: General Management, Entrepreneurship
GMAT 1: 670 Q44 V38 Re: In a certain game only one player can win and only one player always  [#permalink]

Show Tags

EgmatQuantExpert wrote:
sabineodf wrote:
While I am not sure where you are calculating something wrong, I can show you my work process, I am not fantastic at this but I'm sure someone with a little bit more skill than me will have some input haha.

So, since we know that probability of something happening at least once is 1 - (probability of it not happening) we can calculate as follows

In 1 game the probability that James or Austin wins is 3/4th, so the probability that Katelyn wins in 1 game is 1/4th.

They play three games, so she has 3 chances to win: 1/4 + 1/4 + 1/4 = 3/4

Your calculation of 1 - (27/64) I believe is complicating the problem a little bit, I think you multiplied where you should have added. I did that too initially but then I remembered that her playing the game 3 times means her chances of winning increase, not decrease (as they do when you multiply).

I tried to solve this with logic as my math skills aren't great, but there are probably dozens of ways to get the right answer.. Hi sabineodf,

In case of an AND event, we would multiply the probabilities and not add them. The probabilities are added in case its an OR event. To illustrate my point, let me consider the example in the question.

AND Event
For the event where Katelyn wins all 3 games, P(Katelyn winning all 3 games) = P( Katelyn winning the 1st game AND Katelyn winning the 2nd game AND Katelyn winning the 3rd game) $$= \frac{1}{4} * \frac{1}{4} * \frac{1}{4} = \frac{1}{64}$$

In case I extend your logic of adding the probabilities, for 5 games the probability of Katelyn winning would become $$\frac{5}{4}$$, which is not possible.

OR Event
Example of an OR event would be P(Katelyn winning atleast 1 game) = P(Katelyn winning only 1 game) OR P(Katelyn winning only 2 games) OR P(Katelyn winning all 3 games). In this case we would add the probabilities as I have done in the explanation above.

Hope its clear now!

Regards
Harsh

Oooops, thanks! Usually in DS I don't bother working out the math at all, I just ask myself if it can possibly be done with the given informations. Math is my downfall so I tend to make silly mistakes like that Thanks for clearing that up !
Intern  Joined: 13 May 2014
Posts: 11
Re: In a certain game only one player can win and only one player always  [#permalink]

Show Tags

SivaKumarP wrote:
In a certain game only one player can win and only one player always eventually wins. James, Austin, and Katelyn play this game together 3 times in a row. What is the probability that Katelyn wins at least one of the 3 games.?

(1) The probability that either James or Austin wins the game is 3/4
(2) James and Katelyn have an equal probability of winning the game.

Attachment: 1469739715121.jpg [ 78.66 KiB | Viewed 1805 times ]

Sent from my SM-N920R4 using GMAT Club Forum mobile app
Senior Manager  Joined: 02 Mar 2012
Posts: 284
Schools: Schulich '16
Re: In a certain game only one player can win and only one player always  [#permalink]

Show Tags

actually in DS probability question there 's one good thing:

you don't need to calculate the answer.You need to know whether you can arrive at that answer
if this question were a PS one i may have done a silly mistake

1)given probability of other 2 leaving katelyn we can easily calculate the probability of katelyn (1-both)
suff

2)equal probablity

can be 1/2,1/2 or 1/3,/1/3 and many other

insuff

Retired Moderator G
Joined: 11 Aug 2016
Posts: 376
Re: In a certain age only one player can win and one player always eventua  [#permalink]

Show Tags

AkshdeepS wrote:
In a certain game only one player can win and one player always eventually wins. James, Austin, and Katelyn play this game together three times in a row. What is the probability that Katelyn wins at least one of the three games?

1. The probability that either James or Austin wins the game is 3/4.

2. James and Katelyn have an equal probability of winning the game.

Probability that Katelyn wins atleast 1 game=1-(Katelyn wins none of the games)

Statement 1:The probability that either James or Austin wins the game is 3/4.
this means this is the probability of Katelyn loosing a game.
This means we can calculate the required prob.
Sufficient

Statement 2: James and Katelyn have an equal probability of winning the game.
But we dont have the numerical value of their winning the game.
Insufficient.

_________________
~R.
If my post was of any help to you, You can thank me in the form of Kudos!!
Applying to ISB ? Check out the ISB Application Kit. Re: In a certain age only one player can win and one player always eventua   [#permalink] 12 Sep 2018, 11:48
Display posts from previous: Sort by

In a certain game only one player can win and only one player always  