Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If the probability that Mike can win a championship is 1/4, [#permalink]

Show Tags

18 Jan 2008, 17:27

1

This post received KUDOS

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

50% (02:27) correct
50% (03:57) wrong based on 40 sessions

HideShow timer Statictics

If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Rob win the championship but not Ben?

If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Rob win the championship but not Ben?

Probability of Mike OR Rob winning is 1/4 + 1/3 = 7/12 Probability of Ben NOT winning is 6/6-1/6 = 5/6 Probability of Mike OR Rob winning and not Ben is 7/12*5/6 = 35/72

If this is not the right answer, can you pls. explain why this approach doesn't work? Thanks!

If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Rob win the championship but not Ben?

In this situation we are dealing with mutually exclusing events - there is only one person who wins the championship. If Mike or Rob win the championship it automatically means that Ben doesn't win the championship. The answer should be 1/4+1/3 = 7/12.

If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Rob win the championship but not Ben?

In this situation we are dealing with mutually exclusing events - there is only one person who wins the championship. If Mike or Rob win the championship it automatically means that Ben doesn't win the championship. The answer should be 1/4+1/3 = 7/12.

Agree. The same logic. good Q. +1 _________________

In this situation we are dealing with mutually exclusing events - there is only one person who wins the championship. If Mike or Rob win the championship it automatically means that Ben doesn't win the championship. The answer should be 1/4+1/3 = 7/12.

In this situation we are dealing with mutually exclusing events - there is only one person who wins the championship. If Mike or Rob win the championship it automatically means that Ben doesn't win the championship. The answer should be 1/4+1/3 = 7/12.

Agree. The same logic. good Q. +1

Well i thought so to. But apparently the OA is 5/12. The OE is "the probability that Mike or Rob will win is 1/4 + 1/3 or 7/12. The probability that NOT Ben will win is 1/6 and therefore, the total is 7/12 - 1/6 = 5/12". Seems like the OE is replete w/ mistakes.

In this situation we are dealing with mutually exclusing events - there is only one person who wins the championship. If Mike or Rob win the championship it automatically means that Ben doesn't win the championship. The answer should be 1/4+1/3 = 7/12.

Agree. The same logic. good Q. +1

Well i thought so to. But apparently the OA is 5/12. The OE is "the probability that Mike or Rob will win is 1/4 + 1/3 or 7/12. The probability that NOT Ben will win is 1/6 and therefore, the total is 7/12 - 1/6 = 5/12". Seems like the OE is replete w/ mistakes.

Something doesnt make sense here.

There is a problem in the GMATClub challenges that is almost identical to this problem. Except Bens probability not to win is 1/7.

The OA for that one is 1/2. Same process as I did in my above post, so maybe both the OA's are incorrect.

Then the answer is 1/2 and ben's probabilty to lose is 1/7.

Your right. However, the problem you speak of is a replacement of the one i posted. The answer to the previous question (the question in this post) is indeed 7/12 for the reasons Maratikus mentioned.

In this situation we are dealing with mutually exclusing events - there is only one person who wins the championship. If Mike or Rob win the championship it automatically means that Ben doesn't win the championship. The answer should be 1/4+1/3 = 7/12.

Agree. The same logic. good Q. +1

Well i thought so to. But apparently the OA is 5/12. The OE is "the probability that Mike or Rob will win is 1/4 + 1/3 or 7/12. The probability that NOT Ben will win is 1/6 and therefore, the total is 7/12 - 1/6 = 5/12". Seems like the OE is replete w/ mistakes.

In what results should I believe? 7/12 or 5/12. The later is not reliable, because I dont understand the above explained reasoning?

If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Rob win the championship but not Ben?

If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Rob win the championship but not Ben?

Could someone tell what's wrong in this approach? this seems to be make sense.

P=(1/4 *2/3 * 5/6) + (1/3 * 3/4 * 5/6)=25/72 is correct for independent events. For example, we have 3 different game. only one person take part in each game.

P=(1/4 *1 * 1) + (1/3 * 1 * 1)=5/12 is correct for dependent events. For example, we have 1 game. only one person can win. If Mike win the probabilities for Rob and Ben to not win are 1. _________________

If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Rob win the championship but not Ben?

Could someone tell what's wrong in this approach? this seems to be make sense.

P=(1/4 *2/3 * 5/6) + (1/3 * 3/4 * 5/6)=25/72 is correct for independent events. For example, we have 3 different game. only one person take part in each game.

P=(1/4 *1 * 1) + (1/3 * 1 * 1)=5/12 is correct for dependent events. For example, we have 1 game. only one person can win. If Mike win the probabilities for Rob and Ben to not win are 1.[/quote] _____________ You mean 1/4 + 1/3 = 7/12.

If the probability that Mike can win a championship is 1/4, and that of Rob winning is 1/3 and that of Ben winning is 1/6, what is the probability that Mike or Rob win the championship but not Ben?

Soln: Probability that Mike or Bob win is = Prob that Mike will win + prob that Bob will win = 1/4 + 1/3 = 7/12

7/12 and the reasoning seem very logical. However, looking at it from the other end, the probability of 'M' or 'R' winning is the same as that of 'B' losing. Since the probability of B winning is 1/6, the probability of B losing is 1-1/6 = 5/6. This seems pretty logical too

Can somebody please explain why the above approach is wrong?

7/12 and the reasoning seem very logical. However, looking at it from the other end, the probability of 'M' or 'R' winning is the same as that of 'B' losing. Since the probability of B winning is 1/6, the probability of B losing is 1-1/6 = 5/6. This seems pretty logical too

Can somebody please explain why the above approach is wrong?

I think I realized my mistake. My previous argument would have made sense if the sum of the probabilities of M,R and B added to 1. Here, the probability of either one of M, R or B winning is just 3/4 i.e (1/4 + 1/3 + 1/6). So, the probability of none of the guys winning is 1/4.

The probability of B not winning is not the same as M or R winning since this discounts the case where all 3 of them lose.

Hence, the correct answer is (1/4 + 1/3) which amounts to 7/12. Alternatively, it can also be calculated as (3/4 - 1/6) which is also 7/12.

gmatclubot

Re: Probability of Winning
[#permalink]
23 Mar 2010, 04:54

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...