M07-12

Question

If Ben does not win the championship, the probability of Mike winning is \(\frac{1}{4}\), while the probability of Rob winning is \(\frac{1}{3}\). Given that the probability of Ben winning is \(\frac{1}{7}\), what is the probability of either Mike or Rob winning the championship, assuming that only one individual can win?

A. \(\frac{1}{12}\)
B. \(\frac{1}{7}\)
C. \(\frac{1}{2}\)
D. \(\frac{7}{12}\)
E. \(\frac{6}{7}\)

Show Answer

Official Answer

C

A.  \frac{1}{12}  
B.  \frac{1}{7}  
C.  \frac{1}{2}  
D.  \frac{7}{12}  
E.  \frac{6}{7}

Bunuel · Accepted Answer

Official Solution:

If Ben does not win the championship, the probability of Mike winning is \(\frac{1}{4}\), while the probability of Rob winning is \(\frac{1}{3}\). Given that the probability of Ben winning is \(\frac{1}{7}\), what is the probability of either Mike or Rob winning the championship, assuming that only one individual can win?

A. \(\frac{1}{12}\)
B. \(\frac{1}{7}\)
C. \(\frac{1}{2}\)
D. \(\frac{7}{12}\)
E. \(\frac{6}{7}\)

Show Answer

Official Answer

C

A.  \frac{1}{12}  
B.  \frac{1}{7}  
C.  \frac{1}{2}  
D.  \frac{7}{12}  
E.  \frac{6}{7}

This problem involves the concept of conditional probability. It is important to understand that the direct probabilities of Mike and Rob winning the championship are not provided. Instead, we are given the probabilities of Mike and Rob winning the championship, only under the condition that Ben does not win. This means that the probabilities of Mike and Rob winning ( \frac{1}{4}  and  \frac{1}{3} , respectively) are relevant only if we consider the scenario where Ben does not win the championship.
 
The probability of Ben losing is  1 - \frac{1}{7} = \frac{6}{7} . 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 the probability of either Mike or Rob winning the championship is  P = \frac{6}{7}(\frac{1}{4} + \frac{1}{3}) = \frac{1}{2} .
 
Alternatively, consider the following:
 
Take 84 championship cases (84 is chosen as it is the least common multiple of 3, 4, and 7).
 
Out of these 84 cases, Ben will lose in  \frac{6}{7} *84 = 72 . When Ben loses, Mike would win in  72*\frac{1}{4} = 18  cases (which is one-fourth of the cases when Ben loses), and Rob would win in  72*\frac{1}{3} = 24  cases. Therefore, the probability of either Mike or Rob winning the championship is given by  P = \frac{18 + 24}{84} = \frac{1}{2} .

Answer: C

dpgiii · Answer

Hi, I'll take a stab at helping.

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.

Mrtinhnv · Answer

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.

shasadou · Answer

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.

bsattin1 · Answer

samichange

You aren't told that Ben, Mike, and Rob are the ONLY participants. Basically the question could be restated as follows:

If Ben were to lose, Mike would win with probability 1/4, Rob - 1/3, and other - 5/12

Devbek · Answer

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 :evil:

poojashanbhag · Answer

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

Therefore, P(A/B) = 6/7 * 7/12 = 6/12 = 1/2.

bkpolymers1617 · Answer

This is a really nice question. Thanks to the GMAT Club team for providing such questions in Club Tests. Cleared my concepts. Thanks

AnirbanChoudhury89 · Answer

I think this is a  high-quality  question and I agree with explanation.

ArupRS · Answer

Bunuel  and  chetan2u

I did this problem with a scenario LWL+LLW and got an answer which was not in the list. 
Could you please suggest where did I go wrong?

Regards,
Arup

chetan2u · Answer

Hi..

The winning of the other two, M and R, is related to losing of B and nothing else...
When you are taking LWL or LLW, you are making each case, say of M, related to R and B, and that of R, related to M and B..

The question just means that instead of total events as 1, the total events has become 6/7, so scenario is (L of B)(W of M)+(L of B)(W of R) =  \frac{6}{7}*\frac{1}{4}+\frac{6}{7}*\frac{1}{3}=\frac{3}{14}+\frac{2}{7}=\frac{7}{14}=\frac{1}{2}

mSKR · Answer

Mike is winner with probability , Ben lose should be already included in it BECAUSE there can be only one winner.
Similarly ,for Rob would be the winner with a probability of 1/3. If the probability of Ben being the winner is 1/7

Why should it not be 1/4 +1/3 ?
Why i need to consider Ben lose again with probability which is already has winning chances of single person.

I am not able to convince myself  IanStewart  , need your expert opinion. 
thanks!

IanStewart · Answer

I agree that there are issues with the wording of the question, but I think the only reasonable interpretation is the one Bunuel gives in the official solution above. You might imagine a simpler situation:

Serena Williams and Simona Halep will play one tennis match to determine the winner of the 2019 Wimbledon Final. If Williams loses, the probability Halep wins is ____?

The only reasonable answer here is "100%" or 1. We must account for the condition, that Williams loses, when calculating the probability. If the question did not want us to account for that condition, it would not include the phrase "If Williams loses" in the question at all. The same thing is happening in this question, just with a larger group of people competing for the championship.

PyjamaScientist · Answer

AndrewN  I came to this question after reading your post on Conquering Combinatorics last year. But, here I see that the concept of "Conditional Probability" is being used. I have not studied that concept since I learned that Conditional probability isn't something that GMAT tests. So would it be in my interest to go through its theory or give it a pass?

AndrewN · Answer

Hello,  PyjamaScientist . To be clear, for GMAT™ preparation, I would  not  recommend looking up a formula on conditional probability. The question is one that, at its heart, relies on little more than step-by-step logic. Consider the facts:

The probability that Ben will win is 1/7 → The probability that Ben will not win is 6/7 (this sort of reasoning is common in probability questions)

Provided the above holds true, the probability that Mike will win is 1/4; since this meets an  and  condition—i.e. Given #1,  and  #2—we multiply to calculate the probability:

\frac{6}{7} * \frac{1}{4} = \frac{6}{28} = \frac{3}{14}

Repeat the above process to calculate the probability of Rob winning  as an independent event :

\frac{6}{7} * \frac{1}{3} = \frac{6}{21} = \frac{2}{7}

Since we are asked about Mike  or  Rob winning, we add the two independent probabilities:

\frac{3}{14} * \frac{2}{7} = \frac{3}{14} + \frac{4}{14} = \frac{7}{14} = \frac{1}{2}

When something seems complicated on the GMAT™, there is often some simple concept underlying the problem, and that is the case here. If I were to create a somewhat similar question by using a classic standardized test favorite, the six-sided number cube (or what everyone outside of test-taking calls a die or dice, if there are more than one), but with a slight alteration, I think you would hardly toy with the notion that you needed a specific formula, certainly not one on conditional probability, to solve it.

Question: A seven-sided number cube, with the numbers 1 through 7 on each face, respectively, is to be rolled. If a 7 is not rolled, the probability that a 1 will appear face up is 1/6, and a 2, also 1/6. If the number cube is cast but a 7 does not land face up, what is the probability that either a 1 or a 2 will be rolled?

You should hardly have to think about how to solve the question. The only thing that changes in the earlier question is that we are not dealing with a certainty—Ben could win. Do not be afraid to  reason  your way through a question. On a related note, the timer results of some of my own Quant questions have surprised me. (I post one every few months.) I write them with logic in mind, and I think because they are not strictly formulaic, they stump a lot of people. Take a crack at a few of them if you want:

1)  Algebra

2)  Algebra, Divisibility/Multiples/Factors, Roots

3)  Algebra, Coordinate Geometry, Geometry

4)  Divisibility/Multiples/Factors, Number Properties

5)  Arithmetic, Fractions/Ratios/Decimals, Functions and Custom Characters, Inequalities

6)  Algebra, Arithmetic, Exponents/Powers

7)  Arithmetic, Word Problems

8)  Divisibility/Multiples/Factors, Geometry

Have some fun with all this. I have found that in my own preparation experience, I have done my worst when I focus on perfection and on doing everything the  right  way, whatever that means. Well, the right way is the one that gets you the correct answer in a reasonable amount of time, which does not have to be 2 minutes or less. There is almost always room for refinement, but my guess is that if you start approaching Quant questions in more of a mindset that you are aiming to solve a logic puzzle, and you have fun playing Sherlock Holmes, you will actually start to do better. (Just a hunch.)

Thank you for thinking to ask me about this one, and as always, good luck with your studies.

- Andrew

AndrewN · Answer

I can assure you, I did not write any of those questions with traps in mind. Sometimes getting the first domino to fall is the hardest step, after which each subsequent domino gives way with ease. The bell tower question is an example. You have to think,  What do I know about circles and an analog clock face?  Since the problem tells us that it is the top of the hour, we can restrict the options to twelve possibilities. Then, it is a matter of seeing what extra information each statement provides. Perhaps if you were going in order down the list above, your brain was just tired after those other questions. No worries.

One final bit of advice for  this  question: Do not obsess over probability and combinatorics. Just about everybody who has worked in the test prep industry long enough would tell you that students always seem to want to know how to solve these questions in particular, but they are not the most heavily tested of concepts on the exam. A firm foundational knowledge, paired with logic and strong organizational skills, can get you to a 50 in no time, and more advanced questions will not hurt all that much to miss.

- Andrew

PyjamaScientist · Answer

Though I am married, you almost made me fall in love with your words. Man, you should be a motivational speaker more than a tutor!
Thank you for your kind advice. I will try to follow it with all my heart. Thank you for taking out time to write back as always.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

abhinav1239 · Answer

Bunuel  a baisc conceptual doubt. For the case where Mike wins, could you please clarify why you don't need to multiply the probability of Rob not winning e.g. 1/6*1/4*2/3

M07-12

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