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Manager
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Re: M0712
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13 Oct 2017, 07:23
Yeah, that looks like a negative sign there. I understand probability can't be negative, but it did take a few seconds of rereading that. Grammatically, you wouldn't put a dash there anyway. It should be a comma.
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Re: M0712
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20 Dec 2017, 21:59
I think this is a highquality question and I agree with explanation.



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03 Apr 2018, 21:03
I think this the explanation isn't clear enough, please elaborate. As I thought that if of Ben looses the championship that means either Mike or Rob has won so I chose 6/7.



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03 Apr 2018, 21:57
himalayanmonk wrote: I think this the explanation isn't clear enough, please elaborate. As I thought that if of Ben looses the championship that means either Mike or Rob has won so I chose 6/7. You can check more solutions here: https://gmatclub.com/forum/ifbenwere ... 11216.htmlHope it helps.
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Re: M0712
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11 Jul 2018, 20:04
Bunuel wrote: Official Solution:
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. \(\frac{6}{7}\)
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 If Ben were to lose the championship, Mike would be the winner with a probability of \(\frac{1}{4}\), and Rob  \(\frac{1}{3}\) I interpreted the red marked area as Probability(B(not win))*P(M(win))=1/4 Probability(B(not win))*P(R(win))=1/3 Any one pls help me understand the above sentence the significance of "If Ben were to lose the championship"
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11 Jul 2018, 20:53
tejyr wrote: Bunuel wrote: Official Solution:
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. \(\frac{6}{7}\)
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 If Ben were to lose the championship, Mike would be the winner with a probability of \(\frac{1}{4}\), and Rob  \(\frac{1}{3}\) I interpreted the red marked area as Probability(B(not win))*P(M(win))=1/4 Probability(B(not win))*P(R(win))=1/3 Any one pls help me understand the above sentence the significance of "If Ben were to lose the championship" That sentence means that IF Ben loses, then the probability of Mike winning is 1/4 and IF Ben loses, then the probability of Ron winning is 1/3: Mike wining = 6/7*1/4 Rob winning = 6/7*1/3 Mike or Rob wining = 6/7*1/4 + 6/7*1/3
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Re: M0712
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15 Jul 2018, 19:53
Bunuel wrote: tejyr wrote: Bunuel wrote: Official Solution:
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. \(\frac{6}{7}\)
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 If Ben were to lose the championship, Mike would be the winner with a probability of \(\frac{1}{4}\), and Rob  \(\frac{1}{3}\) I interpreted the red marked area as Probability(B(not win))*P(M(win))=1/4 Probability(B(not win))*P(R(win))=1/3 Any one pls help me understand the above sentence the significance of "If Ben were to lose the championship" That sentence means that IF Ben loses, then the probability of Mike winning is 1/4 and IF Ben loses, then the probability of Ron winning is 1/3: Mike wining = 7/6*1/4 Rob winning = 7/6*1/3 Mike or Rob wining = 7/6*1/4 + 7/6*1/3 Hi, Thank you for your reply. I did not get why Mike=7/6*1/4 and why not Mike=6/7*1/4 (Probability(mike win)*Prob(Ben lose)) Can you pls explain.
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Re: M0712
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15 Jul 2018, 20:28
tejyr wrote: Bunuel wrote: tejyr wrote: If Ben were to lose the championship, Mike would be the winner with a probability of \(\frac{1}{4}\), and Rob  \(\frac{1}{3}\) I interpreted the red marked area as Probability(B(not win))*P(M(win))=1/4 Probability(B(not win))*P(R(win))=1/3
Any one pls help me understand the above sentence the significance of "If Ben were to lose the championship" That sentence means that IF Ben loses, then the probability of Mike winning is 1/4 and IF Ben loses, then the probability of Ron winning is 1/3: Mike wining = 7/6*1/4 Rob winning = 7/6*1/3 Mike or Rob wining = 7/6*1/4 + 7/6*1/3 Hi, Thank you for your reply. I did not get why Mike=7/6*1/4 and why not Mike=6/7*1/4 (Probability(mike win)*Prob(Ben lose)) Can you pls explain. It's a typo there. Of course it's 6/7 in both cases. The probability cannot be more than 1.
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Re: M0712
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03 Aug 2018, 10:10
Bunuel Are we not considering the probability(11/3) of Rob losing the game when we are taking the winning probability of Mike since Mike's winning probability already include all the possibilities, even that of Rob losing the game? Is my thinking correct?
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03 Aug 2018, 10:28
Setback wrote: Bunuel Are we not considering the probability(11/3) of Rob losing the game when we are taking the winning probability of Mike since Mike's winning probability already include all the possibilities, even that of Rob losing the game? Is my thinking correct? We are told that there can be only one winner. So, if Mike wins, Rob looses and if Rob wins, Mike looses (automatically).
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Re: M0712
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09 Aug 2018, 13:42
Bunuel wrote: dyg wrote: Hello,
I think question should be edited bcs it seems like Rob's probability of winning is 1/3. Minus should be eliminated from the question. Probability cannot be negative, so it's clearly a dash a not a minus sign. While I am aware that probability cannot be negative, I was briefly thrown off by the dash as well.
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Re: M0712
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04 Sep 2018, 22:50
Hi Bunuel,
I understood the official solution. But I have a doubt in the wording of the question, although I landed up with the correct answer , but I wanted to get it confirmed from you.
According to the question " If Ben were to lose the championship, Mike would be the winner with a probability of 1/4, and Rob  1/3 ", I thought that since only one person can win at at time , P(Mike winning) = 1/4 and P(Rob Losing) = 1/3 are given, so then P(Rob Winning) = 1 P(Rob Losing)= 1(1/3) =2/3 and P(Ben Winning) = 1/7, which is already given . So P(Ben Losing) = 1  P(Ben Winning) = 1(1/7)= 6/7 Now we have to calculate the probability of winning of either Mike or Rob ,
P(Mike Win) = P(Ben Losing)*P(Mike Winning)*P(Rob Losing) = (6/7)*(1/4)*(1/3)=1/14 P(Rob Win) = P(Ben Losing)*P(Mike Losing)*P(Rob Winning) = (6/7)*(3/4)*(2/3)=3/7
P(Mike Win or Rob Win) = P(Mike Win) + P(Rob Win) = (1/14)+(3/7) = 1/2
The final answer is same i.,e 1/2 . But my interpretation was different as compared to official explanation. So I wanted to get it confirmed from you. Whether my approach/understanding is correct?
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Re: M0712
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11 Apr 2019, 19:55
Please change the value for rob from  1/3 to 1/3. It throws off the whole question.







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