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 350,000 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]
26 Jul 2007, 23:38

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 ?

Re: Conceptual Probability Question [#permalink]
27 Jul 2007, 12:04

ajay_gmat wrote:

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 ?

1) 1/6 2) 5/12 3) 7/12 4) 5/6 5) 1

imo, it is

(prob that mike or rob win) (prob that ben won't win) = (1/4 + 1/3) (1 - 1/6) = 35/72

But am going wrong somewhere obviously.. Could you help me out..

they are all exclusive events
(i.e. both people can't win) therefore they are not independant events

Pr A or B = Pr( A) + Pr (B) - Pr( A ANDB)

1/4+1/3=7/12

it might be conceptually simpler to construct two complementary events that sum to 1, PR Ben will win + PR Ben won't win - but that doesn't work in this situation because like others stated we are missing another person.

why are we subtracting Ben's probability of winning?

P(AorB) = P(A) + P(B) - (PAandB)

there can be only one winner so P(AandB)=0

saying P A or B but not C has no meaning because if A or B wins, C can't win.

we are missing another contestant, 'D', with a 3/12 chance of winning... C+D = 5/12.

just seems weird to me.

Yeah I don't really get it... but that's the math rule or something

i hate rules that i don't understand. i will get any problem of this ilk wrong on the gmat, because i don't understand the logic behind it.

this is GMAT prep at its worst. employing a rule/concept you don't understand because you are told it is right.

Just think about it this way:
The total probability at least ONE of them wins:
P(Mike Wins) + P(Rob Wins) + P(Ben Wins) = 9/12

If they told you that you could not count on Ben winning, so now what are the chances of total win between Mike and Rob with Ben's loss?
-- Here you're subtracting the possibility of Ben winning all together from their combined probability. So:
P(Mike) + P(Rob) - P(Ben) = 7/12

Social entrepreneurs aren't running charities : ‘It’s easy to think about it in terms of charity: it’s not. They are sustainable, trading, revenue-generating businesses. The benefit of them...