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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

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An (x, y) coordinate pair is to be chosen at random from the [#permalink]
20 Dec 2014, 05:45

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 _________________

+1 Kudos if my comment was helpful. Thanks!

Failure forges confidence, and confidence cultivates success. Proving the answer choices wrong is almost better than calculating what is right.

gmatclubot

An (x, y) coordinate pair is to be chosen at random from the
[#permalink]
30 Jun 2015, 18:12

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