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:
The y is positive in two quadrants - so \(P[y\geq0]=\frac{1}{2}\).
(I am deliberately ignoring here y=0, because theoretically the axes lie inbetween the quadrants, so 0 isn't part of any quadrant equally.)
So, when \(P[y\geq0=\frac{1}{2}]\), we need to figure out \(P[y>x]\).
Since there is infinite number of possibilities for positive values of x and y, I concluded that there's a 50% chance of picking (x, y) at random, such that \(y>x\).
(Here's another apparent leave-out: the possibility that \(x=y\). However, this is only one possibility among infinity, so as for me it's acceptable. Please correct me if I am wrong.)
This way we've got \(P[y\geq0]=\frac{1}{2}\) and \(P[y>x]=\frac{1}{2}\).
The y is positive in two quadrants - so \(P[y\geq0]=\frac{1}{2}\).
(I am deliberately ignoring here y=0, because theoretically the axes lie inbetween the quadrants, so 0 isn't part of any quadrant equally.)
So, when \(P[y\geq0=\frac{1}{2}]\), we need to figure out \(P[y>x]\).
Since there is infinite number of possibilities for positive values of x and y, I concluded that there's a 50% chance of picking (x, y) at random, such that \(y>x\).
(Here's another apparent leave-out: the possibility that \(x=y\). However, this is only one possibility among infinity, so as for me it's acceptable. Please correct me if I am wrong.)
This way we've got \(P[y\geq0]=\frac{1}{2}\) and \(P[y>x]=\frac{1}{2}\).
Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink]
09 Jun 2014, 07:05
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
An (x, y) coordinate pair is to be chosen at random from the [#permalink]
20 Dec 2014, 05:45
1
This post received KUDOS
What I did -->
We know that we have to choose a point such that Y is positive--> so while choosing any number from the number line..the probability of selecting a number >0 or <0 should be equal..So Prob of choosing a positive Y is 1/2. Say a coordinate x is chosen...Now,The probability of choosing any number that is <x or >x is again equal..so the Probability is again 1/2 for each case.
Now for P(y) such that both the conditions are met will be 1/2*1/2= 1/4.
Can any expert vouch for this way of solving such a question?? _________________
Appreciate the efforts...KUDOS for all Don't let an extra chromosome get you down..
An (x, y) coordinate pair is to be chosen at random from the [#permalink]
04 Jan 2015, 02:10
JusTLucK04 wrote:
What I did -->
We know that we have to choose a point such that Y is positive--> so while choosing any number from the number line..the probability of selecting a number >0 or <0 should be equal..So Prob of choosing a positive Y is 1/2. Say a coordinate x is chosen...Now,The probability of choosing any number that is <x or >x is again equal..so the Probability is again 1/2 for each case.
Now for P(y) such that both the conditions are met will be 1/2*1/2= 1/4.
Can any expert vouch for this way of solving such a question??
Now for P(y) such that both the conditions are met will be 1/2*1/2= 1/4. this is simply wrong as you are reading P(y>=|x|) as P(y>|x| AND y=|x|) this should be 0. question is asking P(y>=|x|) which is P(y>|x| ) OR P(y=|x|) -> P(y>|x| ) + P(y=|x|) _________________
Thanks, Lucky
_______________________________________________________ Kindly press the to appreciate my post !!
An (x, y) coordinate pair is to be chosen at random from the [#permalink]
30 Jun 2015, 18:12
JusTLucK04 wrote:
What I did -->
We know that we have to choose a point such that Y is positive--> so while choosing any number from the number line..the probability of selecting a number >0 or <0 should be equal..So Prob of choosing a positive Y is 1/2. Say a coordinate x is chosen...Now,The probability of choosing any number that is <x or >x is again equal..so the Probability is again 1/2 for each case.
Now for P(y) such that both the conditions are met will be 1/2*1/2= 1/4.
Can any expert vouch for this way of solving such a question??
Had there been constraints, such as \(y\geq{|\frac{1}{2}x|}\), or say \(y\geq{x^{2}}\), the range would vary enough to where \(\frac{1}{2} * \frac{1}{2}\) estimation would be inaccurate, and therefore the solution above would not have worked. You might have adjusted for it, but since the question is out there, I took a swipe. Cheers
gmatclubot
An (x, y) coordinate pair is to be chosen at random from the
[#permalink]
30 Jun 2015, 18:12
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...