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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. \(\frac{6}{7}\)
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 \(P=\frac{18+24}{84}=\frac{1}{2}\).
Dear Bunuel, Here is my solution, P(Mike win) = 6/7*1/4*2/3 P(Rob win) = 6/7*1/3*3/4 => P(Mike win or Rob win) = 6/7*1/4*2/3 + 6/7*1/3*3/4 = 5/14 How is my solution wrong? I don't understand why P=6/7(1/4+1/3), The question is either Mike or Rob will win Please help me.
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
Hi Bunnel
If 6/7 * 1/4 = Probability that Mike Wins, Ben Loses and Rob Loses then 6/7 * 3/4 = Probability that Mike loses, Ben Loses and Rob WINS
( Given that there can only be ONE Winner and NO Draw)
BUT the probability that Mikes loses, Ben Loses and Rob Wins = 6/7 * 1/3
Where am I wrong?Probability is definitely not my forte
Please clarify
Thanks
_________________
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I hated every minute of training, but I said, 'Don't quit. Suffer now and live the rest of your life as a champion.-Mohammad Ali
Dear Bunuel, Here is my solution, P(Mike win) = 6/7*1/4*2/3 P(Rob win) = 6/7*1/3*3/4 => P(Mike win or Rob win) = 6/7*1/4*2/3 + 6/7*1/3*3/4 = 5/14 How is my solution wrong? I don't understand why P=6/7(1/4+1/3), The question is either Mike or Rob will win Please help me.
the stem says "If Ben were to lose the championship, Mike would be the winner with a probability of 1/4, and Rob - 1/3" - that is those probabilities of mike and rob are dependent on Ben losing and are not dependent on each other: whether mike loses then rob should win or vice versa. this is also what confused me.
so given that the stem gives us the OR probability of two events dependent on Ben only - we should add up 1/3 and 1/4. we cannot solve for scenarios when mike's win depends on the lose of rob because stem does not give this therefore 6/7*1/4*2/3 is not correct (we cannot make the assumption).
on the other hand if the stem asked to find the individual probability of winning of either mike or rob then the stem should ahve provided some more info on the dependency.
_________________
Dear Bunuel, Here is my solution, P(Mike win) = 6/7*1/4*2/3 P(Rob win) = 6/7*1/3*3/4 => P(Mike win or Rob win) = 6/7*1/4*2/3 + 6/7*1/3*3/4 = 5/14 How is my solution wrong? I don't understand why P=6/7(1/4+1/3), The question is either Mike or Rob will win Please help me.
Hi, I had the same solution. I don't know why other's loss wasn't considered since the question stem says only one winner is possible.
As mentioned above, the probability that Ben wins the championship is 1/7. Therefore, the probability that Ben loses the championship is 6/7 (in other words, 1- 1/7 = 6/7).
Mike has a 1/4 chance of winning. This is conditional upon Ben losing (because Ben and Mike can't both win). To calculate Mike's chances of winning, you multiply the probability that Mike wins (1/4) * the probability that Ben loses (6/7).
1/4 * 6/7 = 3/14
Repeat the same process for Rob. Multiply the probability of Ben losing times the probability of Rob winning
1/3 * 6/7 = 2/7
Simply add the two probabilities together. You add the probabilities together because the question asks what the probability that Mike OR Rob win.
The result is 3/14 + 2/7 = 7/14 = 1/2.
As Bunuel would say, Hope this helps.
Bunuel 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. \(\frac{6}{7}\)
I think this question is good and helpful. Please edit the question prompt. It looks like Rob's probability of winning is -(1/3)
It's simply a dash and not the negative sign. Probability of an occurrence can't be negative. If the probability of something occurring is 0, then it is impossible for it to occur. Probability cannot be any lower than impossible.
The tricky part is whether two have scenarios with three people or not. I first did it with a scenario LWL+LLW and got an answer which was not in the list. Then I understood my mistake which is making my own assumption
I think this is a high-quality question and I agree with explanation. one thing is that -1\3 looks like negative. it makes me confuse.
_________________
The probability of an event is always between 0 and 1, inclusive. So the question of some probability in the question given as -1/3 doesn't arise.
Secondly, probability of an event given that another event has occurred is P(A/B) = P(A)*P(B)
Here, P(A/B) is chances of other guys winning given that Ben has lost P(A) = Ben has lost ... which has been derived as 1 - P(A') = 1 - 1/7 where P(A') = probability that Ben wins. P(B) = Other 2 guys winning = 1/4+ 1/3 = 7/12
Whether this can be a way of answering this question.
Probability(Mike or Rob will win)=Probability of Mike winning+Probability of Rob winning-(Probability of Mike and Rob both winning) =1/4 +1/3 -1/4*1/3(Since its already mentioned ,these are the probabilities subject to Ben losing so Ben's winning is anyways not required) =1/2(Answer)
This is a really nice question. Thanks to the GMAT Club team for providing such questions in Club Tests. Cleared my concepts. Thanks
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