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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]  18 Jan 2013, 16:38
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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 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]  18 Jan 2013, 17:04
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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 correct and I hope it's not pure luck .

Below is given graph of y=|x|:
Attachment:

Graph.png [ 11.52 KiB | Viewed 514 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.

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Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink]  18 Jan 2013, 17:12
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KUDOS
Attachment:

XY Plane probability.jpg [ 171.15 KiB | Viewed 518 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

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PraPon

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Re: An (x, y) coordinate pair is to be chosen at random from the   [#permalink] 18 Jan 2013, 17:12
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