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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 _________________

Kr, mejia401

+1 Kudos if my comment was helpful. Thanks! Failure forges confidence, confidence multiplies success.

gmatclubot

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

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