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An (x, y) coordinate pair is to be chosen at random from the

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An (x, y) coordinate pair is to be chosen at random from the [#permalink] New post 18 Jan 2013, 15:38
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An (x, y) coordinate pair is to be chosen at random from the xy-plane. What is the probability that \(y\geq|x|\)?

(A) 1/10
(B) 1/8
(C) 1/6
(D) 1/5
(E) 1/4


Here's how I bumped on the answer - very quickly, but I am not 100% sure if logic I used was correct. Someone could proof please?
[Reveal] Spoiler:
The |x| is always positive.

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 \(P[y>x]=\frac{1}{2}\).

Therefore \(P[y\geq|x|]=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}\).

The answer is E.

The answer is correct and I hope it's not pure luck :).
[Reveal] Spoiler: OA

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Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink] New post 18 Jan 2013, 16:04
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HumptyDumpty wrote:
An (x, y) coordinate pair is to be chosen at random from the xy-plane. What is the probability that \(y\geq|x|\)?

(A) 1/10
(B) 1/8
(C) 1/6
(D) 1/5
(E) 1/4


Here's how I bumped on the answer - very quickly, but I am not 100% sure if logic I used was correct. Someone could proof please?
[Reveal] Spoiler:
The |x| is always positive.

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 \(P[y>x]=\frac{1}{2}\).

Therefore \(P[y\geq|x|]=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}\).

The answer is E.

The answer is correct and I hope it's not pure luck :).


Below is given graph of y=|x|:
Attachment:
Graph.png
Graph.png [ 11.52 KiB | Viewed 2829 times ]
All points which satisfy \(y\geq|x|\) condition lie above that graph. You can see that portion of the plane which is above the graph is 1/4.

Answer: E.
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Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink] New post 18 Jan 2013, 16:12
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Attachment:
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XY Plane probability.jpg [ 171.15 KiB | Viewed 2831 times ]


Just plot the y>|x| on your paper and you will see it takes 1/4 of the space i.e. Probability = 1/4

Hence choice(E) is the answer.
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Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink] New post 09 Jun 2014, 07:05
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An (x, y) coordinate pair is to be chosen at random from the [#permalink] New post 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??
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An (x, y) coordinate pair is to be chosen at random from the [#permalink] New post 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|)
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An (x, y) coordinate pair is to be chosen at random from the [#permalink] New post 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
An (x, y) coordinate pair is to be chosen at random from the   [#permalink] 30 Jun 2015, 18:12
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