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

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In a certain game only one player can win and only one player always  [#permalink]

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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, 18:37.
Last edited by Bunuel on 24 Apr 2015, 01:28, edited 1 time in total.
Renamed the topic and edited the question.
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Re: In a certain game only one player can win and only one player always  [#permalink]

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New post 23 Apr 2015, 18:49
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.

Answer is A
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Re: In a certain game only one player can win and only one player always  [#permalink]

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New post 23 Apr 2015, 18:59
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 :shock:
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Re: In a certain game only one player can win and only one player always  [#permalink]

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New post 23 Apr 2015, 19:19
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.. :-D
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Re: In a certain game only one player can win and only one player always  [#permalink]

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New post 24 Apr 2015, 01:01
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. :shock:
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Re: In a certain game only one player can win and only one player always  [#permalink]

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New post 24 Apr 2015, 01:09
2
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 :shock:


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
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Re: In a certain game only one player can win  [#permalink]

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New post 24 Apr 2015, 01:31
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.. :-D


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
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Re: In a certain game only one player can win and only one player always  [#permalink]

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New post 24 Apr 2015, 02:14
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 :shock:


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
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Re: In a certain game only one player can win and only one player always  [#permalink]

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New post 24 Apr 2015, 03:25
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 :shock:


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
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Re: In a certain game only one player can win and only one player always  [#permalink]

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New post 24 Apr 2015, 07:29
Thanks Harsh. I got it and I will try to become master with that :lol:

This time instead of saying kudos, I really gave you kudos.. :-D
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Re: In a certain game only one player can win and only one player always  [#permalink]

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New post 24 Apr 2015, 18:31
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.. :-D


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 :shock: Thanks for clearing that up !
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Re: In a certain game only one player can win and only one player always  [#permalink]

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New post 28 Jul 2016, 13:02
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.

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Re: In a certain game only one player can win and only one player always  [#permalink]

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New post 29 Jul 2016, 00:08
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

here the answer is A

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


answer is A
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Re: In a certain age only one player can win and one player always eventua  [#permalink]

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New post 12 Sep 2018, 10:48
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.

Answer: A
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Re: In a certain age only one player can win and one player always eventua &nbs [#permalink] 12 Sep 2018, 10:48
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