Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 19 Sep 2010
Posts: 21

If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
02 Oct 2010, 05:57
Question Stats:
45% (02:09) correct 55% (02:33) wrong based on 96 sessions
HideShow timer Statistics
If Ben were to lose the championship, Mike would be the winner with a probability of \(\frac{1}{4}\), and Rob  \(\frac{1}{3}\) . If the probability of Ben being the winner is \(\frac{1}{7}\), what is the probability that either Mike or Rob will win the championship? Assume that there can be only one winner. A. \(\frac{1}{12}\) B. \(\frac{1}{7}\) C. \(\frac{1}{2}\) D. \(\frac{7}{12}\) E. \(\frac{6}{7}\) M0712
Official Answer and Stats are available only to registered users. Register/ Login.



Intern
Joined: 19 Sep 2010
Posts: 21

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
02 Oct 2010, 06:00
The OA : \((1\frac{1}{7}) * \frac{7}{12} = \frac{6}{7} * \frac{7}{12} = \frac{1}{2}\) . (7/12 = 1/3 + 1/4)
Can Some one explain me what is the problem with this reasoning : P=(1/4 *2/3 * 6/7) + (1/3 * 3/4 * 6/7)= 5/14
Thanks



Retired Moderator
Joined: 02 Sep 2010
Posts: 711
Location: London

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
02 Oct 2010, 07:12
Barkatis wrote: The OA : \((1\frac{1}{7}) * \frac{7}{12} = \frac{6}{7} * \frac{7}{12} = \frac{1}{2}\) . (7/12 = 1/3 + 1/4)
Can Some one explain me what is the problem with this reasoning : P=(1/4 *2/3 * 6/7) + (1/3 * 3/4 * 6/7)= 5/14
Thanks I think what you have done is : Probability = P(Ben loses) * P(Mike wins given Ben loses) * (1  P(Rob wins givne Ben loses)) + P(Ben loses) * P(Rob wins given Ben loses) * (1  P(Mike wins givne Ben loses)) The problem is the terms I have marked in red. You are already given the probability that Mike wins once Ben has lost, 1/4, you do not need to multiply this with (12/3). Similarly in the second term The correct answer would be : P(Ben loses) * P(Mike wins given Ben loses) + P(Ben loses) * P(Rob wins given Ben loses) OR (6/7) * (1/3) + (6/7) * (1/4) = (1/2)
_________________



Intern
Joined: 19 Sep 2010
Posts: 21

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
02 Oct 2010, 12:11
Thanks for answering.
I got that part but my question is how do we know that the probability given is that Mike wins ONCE BEN HAS LOST.
What is the difference between that problem and this one for example Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4 ,1/2 , and 5/8 , respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem ?
Where the solution is 3/8*1/2*1/4 = 3/64 But if we assume that the probability given are those of success of one person while the two others are loosing it would be more : 1/4*1/2 . Right ?
I hope you got my problem better now.



Retired Moderator
Joined: 02 Sep 2010
Posts: 711
Location: London

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
02 Oct 2010, 17:26
Barkatis wrote: Thanks for answering.
I got that part but my question is how do we know that the probability given is that Mike wins ONCE BEN HAS LOST.
What is the difference between that problem and this one for example Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4 ,1/2 , and 5/8 , respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem ?
Where the solution is 3/8*1/2*1/4 = 3/64 But if we assume that the probability given are those of success of one person while the two others are loosing it would be more : 1/4*1/2 . Right ?
I hope you got my problem better now. You have to be careful about reading the wording. In the example you are giving, there are 3 people solving a question. None, one or more could solve it correctly. In the question at hand, there are 3 players trying to win a tournament. Either none of them or atmost one of them can win, and simultaneous winning is not possible. Now whether we have the probability of Mike winning given Ben has lost or we have absolute probability of Mike winning is just how the question is worded. If Ben were to lose the championship, Mike would be the winner with a probability of \frac{1}{4}
_________________



Manager
Joined: 05 Jan 2011
Posts: 92

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
21 Mar 2011, 07:08
asmit123 wrote: If Ben were to lose the championship,MIke would be the winner with a probability of 1/4, and Rob  1/3. If probab of Ben winning is 1/7, what is the probab that either Mike or Rob will win the championship(there can be only one winner) ?
1.1/12 2.1/7 3.1/2 4.7/12 5.6/7 Probability Ben will lose =11/7 =>6/7 Probability Ben will lose and Mike will win => 1/4*6/7 Probability Ben will lose and Rob will win=>1/3*6/7 Probability Either Mike or Rob will win=> 6/7(1/4+1/3) 6/7 *7/12=> 1/2 C



Director
Status: =Given to Fly=
Joined: 04 Jan 2011
Posts: 784
Location: India
Concentration: Leadership, Strategy
GMAT 1: 650 Q44 V37 GMAT 2: 710 Q48 V40 GMAT 3: 750 Q51 V40
GPA: 3.5
WE: Education (Education)

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
21 Mar 2011, 07:09
Ans = Prob. Ben will lose and Mike will win OR Prob. Ben will lose and Rob will win = 6/7x1/4 + 6/7x1/3 =6/7 x (1/4+1/3) =6/7 x (7/12) =1/2
_________________



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 16107
Location: United States (CA)

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
25 Jan 2015, 13:08
Hi 23a2012, This question is based on "conditional" probability, which is an exceptionally rare subject on the GMAT (you probably will NOT see it at all on Test Day  if you do, then it would likely be in a DS question). Unless you're already scoring in the Q49+ range, you be spending your time on other subjects. GMAT assassins aren't born, they're made, Rich
_________________
Contact Rich at: Rich.C@empowergmat.comThe Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★



Math Expert
Joined: 02 Sep 2009
Posts: 61283

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
26 Jan 2015, 01:57
asmit123 wrote: If Ben were to lose the championship, Mike would be the winner with a probability of \(\frac{1}{4}\), and Rob  \(\frac{1}{3}\) . If the probability of Ben being the winner is \(\frac{1}{7}\), what is the probability that either Mike or Rob will win the championship? Assume that there can be only one winner.
A. \(\frac{1}{12}\) B. \(\frac{1}{7}\) C. \(\frac{1}{2}\) D. \(\frac{7}{12}\) E. \(\frac{6}{7}\)
M0712 This is a conditional probability question. We need the probability that either Mike or Rob will win the championship. So Ben must lose: the probability of Ben losing is \(1\frac{1}{7}=\frac{6}{7}\). Now out of these \(\frac{6}{7}\) cases the probability of Mike winning is \(\frac{1}{4}\) and the probability of Rob winning is \(\frac{1}{3}\). So \(P=\frac{6}{7}(\frac{1}{4}+\frac{1}{3})=\frac{1}{2}\). Or consider the following: Take 84 championships/cases (I chose 84 as it's a LCM of 3, 4, and 7). Now, out of these 84 cases Ben will lose in \(\frac{6}{7}*84=72\). Mike would be the winner in \(72*\frac{1}{4}=18\) (1/4 th of the cases when Ben loses) and Rob would be the winner in \(72*\frac{1}{3}=24\). Therefore \(P=\frac{18+24}{84}=\frac{1}{2}\). Answer: C.
_________________



Intern
Joined: 16 May 2015
Posts: 28

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
15 Jul 2015, 22:26
Hi All,
I am confused..i thought the probability that Ben will lose = P of either Mike r Rob will win? or we are assuming that there are still others in the championship? so my earlier assumption is not valid?
Thanks in advance



Senior Manager
Joined: 28 Jun 2015
Posts: 274
Concentration: Finance
GPA: 3.5

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
15 Jul 2015, 22:52
Conditional probability question. Probability of Ben winning = 1/7; so probability of Ben losing = 6/7. Probability of Mike winning given that Ben had lost = 1/4 * 6/7 = 6/28 = 3/14 Probability of Rob winning given that Ben had lost = 1/3 * 6/7 = 6/21. = 2/7 = 4/14 Required probability = 3/14 + 4/14 = 7/14 = 1/2. Ans (C).
_________________
I used to think the brain was the most important organ. Then I thought, look what’s telling me that.



Senior Manager
Joined: 28 Jun 2015
Posts: 274
Concentration: Finance
GPA: 3.5

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
15 Jul 2015, 23:00
katzzzz wrote: Hi All,
I am confused..i thought the probability that Ben will lose = P of either Mike r Rob will win? or we are assuming that there are still others in the championship? so my earlier assumption is not valid?
Thanks in advance Hi, I don't quite understand your doubt, see if this helps... Probability that Ben wins = 1/7 (given); so the 'counterevent' would be Ben losing = 1  1/7 = 6/7. Now, they have asked to find the probability that EITHER Rob or Mike wins, and there are 2 conditions imposed: (1) there can only be one winner (2) (stemming from the 1st condition) Ben loses (because we have to find only the probability of Mike and Rob, which means Ben loses as there can be only one winner). Rob winning the game given that Ben had lost = 1/3 * 6/7 Mike winning given that Ben had lost = 1/4 * 6/7 Adding those two we get the answer.
_________________
I used to think the brain was the most important organ. Then I thought, look what’s telling me that.



Intern
Joined: 16 May 2015
Posts: 28

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
16 Jul 2015, 02:00
Hi, I mean..why can't we choose 6/7 as the answer. Probability of either one of them win = probability of mike lost the game? I not sure where did get it wrong.. Thanks



Math Expert
Joined: 02 Aug 2009
Posts: 8261

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
16 Jul 2015, 02:23
katzzzz wrote: Hi, I mean..why can't we choose 6/7 as the answer. Probability of either one of them win = probability of mike lost the game? I not sure where did get it wrong.. Thanks hi, the probability given for other two to be winner shows that there are other in the race.. otherwise the total probability should have been 1 for all three winning ..
_________________



Senior Manager
Joined: 28 Jun 2015
Posts: 274
Concentration: Finance
GPA: 3.5

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
16 Jul 2015, 03:13
katzzzz wrote: Hi, I mean..why can't we choose 6/7 as the answer. Probability of either one of them win = probability of mike lost the game? I not sure where did get it wrong.. Thanks The most simple and direct answer to that question is the events are independent. I hope you get it now?
_________________
I used to think the brain was the most important organ. Then I thought, look what’s telling me that.



Intern
Joined: 14 Apr 2015
Posts: 14

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
18 Jul 2015, 10:03
FireStorm wrote: katzzzz wrote: Hi, I mean..why can't we choose 6/7 as the answer. Probability of either one of them win = probability of mike lost the game? I not sure where did get it wrong.. Thanks The most simple and direct answer to that question is the events are independent. I hope you get it now? i thought that probability of mike winning =(Prob. of Ben not winning)*(Prob. of Rob not winning)*(Prob. of Mike winning) FRom all explanations above it seems that championship does not have only 3 players But how can i identify this ? Thanks



Senior Manager
Joined: 28 Jun 2015
Posts: 274
Concentration: Finance
GPA: 3.5

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
18 Jul 2015, 18:12
divya517 wrote: FireStorm wrote: katzzzz wrote: Hi, I mean..why can't we choose 6/7 as the answer. Probability of either one of them win = probability of mike lost the game? I not sure where did get it wrong.. Thanks The most simple and direct answer to that question is the events are independent. I hope you get it now? i thought that probability of mike winning =(Prob. of Ben not winning)*(Prob. of Rob not winning)*(Prob. of Mike winning) FRom all explanations above it seems that championship does not have only 3 players But how can i identify this ? Thanks Hi, You are looking at the wrong set of events. There may or may not be other players, but we are asked to find out the probability of either Mike or Rob winning with respect to the probability of Ben losing, which is why we don't go about solving this the way you did [probability of mike winning =(Prob. of Ben not winning)*(Prob. of Rob not winning)*(Prob. of Mike winning)]. Hope it's clear.
_________________
I used to think the brain was the most important organ. Then I thought, look what’s telling me that.



Math Expert
Joined: 02 Aug 2009
Posts: 8261

If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
18 Jul 2015, 20:35
FireStorm wrote: divya517 wrote: FireStorm wrote: The most simple and direct answer to that question is the events are independent. I hope you get it now?
i thought that probability of mike winning =(Prob. of Ben not winning)*(Prob. of Rob not winning)*(Prob. of Mike winning) FRom all explanations above it seems that championship does not have only 3 players But how can i identify this ? Thanks Hi, You are looking at the wrong set of events. There may or may not be other players, but we are asked to find out the probability of either Mike or Rob winning with respect to the probability of Ben losing, which is why we don't go about solving this the way you did [probability of mike winning =(Prob. of Ben not winning)*(Prob. of Rob not winning)*(Prob. of Mike winning)]. Hope it's clear. Hi, there have to be more players... otherwise divya517 & katzzzz are correct , the answer would be 6/7... But the Question is giving probability of the two player winning when ben does not win, and since the combined prob of either of them winning is not equal to ben losing, so there are more player who too can win if ben loses
_________________



Senior Manager
Joined: 28 Jun 2015
Posts: 274
Concentration: Finance
GPA: 3.5

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
18 Jul 2015, 22:25
Hi, Yes, there are more. But strictly speaking that's a circuitous route of going around this problem. We are asked to find the probability only w.r.t Ben's chances of winning/losing.
_________________
I used to think the brain was the most important organ. Then I thought, look what’s telling me that.



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 3151
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
Show Tags
18 Jul 2015, 22:37
asmit123 wrote: If Ben were to lose the championship, Mike would be the winner with a probability of \(\frac{1}{4}\), and Rob  \(\frac{1}{3}\) . If the probability of Ben being the winner is \(\frac{1}{7}\), what is the probability that either Mike or Rob will win the championship? Assume that there can be only one winner.
A. \(\frac{1}{12}\) B. \(\frac{1}{7}\) C. \(\frac{1}{2}\) D. \(\frac{7}{12}\) E. \(\frac{6}{7}\)
M0712 Case1: Ben loses and Mile Wins The probability of Ben losing is 6/7 and then Mike winning is 1/4 so the total Probability of Case1 = (6/7)*(1/4) Case2: Ben loses and Mile Wins The probability of Ben losing is 6/7 and then Rob winning is 1/3 so the total Probability of Case1 = (6/7)*(1/3) Total Probability of Case1&2 together = (6/7)*(1/4)+(6/7)*(1/3) = 1/2 Answer: Option C
_________________
Prosper!!!GMATinsightBhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhihttp://www.GMATinsight.com/testimonials.htmlClick here for Our VERBAL & QUANT private tutoring package detailsACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION




Re: If Ben were to lose the championship, Mike would be the winner with a
[#permalink]
18 Jul 2015, 22:37



Go to page
1 2
Next
[ 22 posts ]



