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 Post subject: GMAT Coordinate Geometry
PostPosted: Sat Jul 10, 2010 5:24 am 
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The center of a circle lies on the origin of the coordinate plane. If a point (x, y) is randomly selected inside of the circle, what is the probability that x > y > 0?

A. 1/8
B. 1/6
C. 3/8
D. 1/2
E. 3/4

(A) Since the circle is centered on the origin, exactly 1/4 of the points in the circle will consist of points where both x and y are positive (those in Region I in the coordinate plane).

Of these points, half will lie above the line y = x (where y > x) and half will lie below it (where x > y).

Thus, 1/8 of the circle will consist of all the points in which x > y > 0.

The correct answer is choice (A).
-------------

Quadrant 1: 1/2 chance x > y
Quadrant 2: x > y (as y is negative while x is positive)
Quadrant 3: 1/2 chance x > y
Quadrant 4: x < y (as y is positive while x is negative)

Therefore, answer should be 1/2*1/2 + 1/2 + 0 + 1/2*1/2 = 7/8.

Please advice. Thanks a lot!


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 Post subject: Re: math (test 1, question 26): probability, x,y-plane
PostPosted: Sat Jul 10, 2010 6:11 am 
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Be very careful when dealing with probability questions, and watch closely if you have the same probabilistic event all the way or you've changed it and need to make corresponding adjustments.

First, I'll show the most general approach, which works very well in this case. Then, I'll analyze yours and give you some advices.

The most general approach when finding probability in geometrical questions is to use definition of probability on plane:

The probability equals to the area of the figure that consists of satisfactory probabilistic events (In our case it consists of all the points within a circle such that x > y > 0) divided by the area of the figure that consists of all the possible probabilistic events (In our case we pick a point only within circle so it is a circle itself except circumference).

Let radius of the circle be r.
All the possible events:
Image
Area equals πr²

Satisfactory events:
Image
Area equals (1/8) × πr²

So the probability is [(1/8) × πr²] / (πr²) = 1/8.

Your approach is also possible but it is more complicated. let me analyze it:
When you calculate probability in each quadrant separately you change probabilistic event by picking a point not from a circle at whole but just from 1/4 of it that lies in the corresponding quadrant.
So we need to adjust the event by assuming that we pick one of the quadrants first (equiprobably, because the parts of the circle we pick, 1/4 of a circle, all have the same area). And the probability we need to calculate equals to:
(1/4) × P1 + (1/4) × P2 + (1/4) × P3 + (1/4) × P4,
where P1, P2, P3 and P4 probabilities in each quadrant correspondingly.
Let us simplify:
1/4 (P1 + P2 + P3 + P4)

Now, note that x > y > 0. So all P2, P3 and P4 are 0. P1, as you've calculated it right, is 1/2. So the answer is:
1/4 (1/2 + 0 + 0 + 0) = 1/8.

Furthermore, if we broke the figure into different areas, like 1/4, 1/8, 1/8, 1/2 of a circle, for example, we would count probability as
(1/4) × P1 + (1/8) × P2 + (1/8) × P3 + (1/2) × P4.


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 Post subject: Re: math (test 1, question 26): probability, x,y-plane
PostPosted: Sun Nov 28, 2010 5:26 pm 
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I don't think it's that simple... what about the points in which x = y? (e.g. 2,2). It's not always the case that x is greater than or less than y.


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 Post subject: Re: math (test 1, question 26): probability, x,y-plane
PostPosted: Sun Nov 28, 2010 5:37 pm 
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When we deal with probability on the x,y-plane and the set of possible probabilistic events has certain area then we deal with areas.
probability = suitable area / total possible area

So in this case figures that do not have area, do NOT affect the result, like lines, points etc.

Note, that when the set of possible probabilistic events is:
- line, then we deal with segments
- finite set of points, then we deal with regular (discrete) definition of probability (see the following problem as an example: http://800score.com/forum/viewtopic.php?t=67)


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 Post subject: Re: math (test 1, question 26): probability, x,y-plane
PostPosted: Sun Aug 28, 2011 3:47 pm 
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How do we get 1/8 for the area of shaded region?

Thanks in Advance!


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 Post subject: Re: math (test 1, question 26): probability, x,y-plane
PostPosted: Sun Aug 28, 2011 4:39 pm 
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The x-axis and y-axis divide the circle into quarters (1/4).

The line y = x divides the 90⁰-angle (between the coordinate axes) in half, into two 45⁰-angles. 1/2 from 1/4 is 1/2 × 1/4 = 1/8.


Alternative way. The angle of the shaded sector of the circle is 45⁰. Therefore its area is 45⁰/360⁰ = 1/8.
Image


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 Post subject: Re: math (test 1, question 26): probability, x,y-plane
PostPosted: Tue Oct 11, 2011 11:33 am 
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In this question, I understand that Quadrant 1 has all the required values. Is the line y = x a slant line rising upwards to the right from the origin?


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 Post subject: Re: math (test 1, question 26): probability, x,y-plane
PostPosted: Tue Oct 11, 2011 11:36 am 
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Quote:
Is the line y = x a slant line rising upwards to the right from the origin?
In order to draw a line we need to have two points.

If x = 0 then y = 0. This gives us point (0, 0).
If x = 1 then y = 1. This gives us point (1, 1).

You may look at the line on the pictures above.


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 Post subject: Re: math (test 1, question 26): probability, x,y-plane
PostPosted: Wed Nov 16, 2011 11:01 am 
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Why are we considering only the first quadrant? The question does not mention anything about x & y being positive only. In the 4th quadrant, wont all the values of x be greater than that of y?


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 Post subject: Re: math (test 1, question 26): probability, x,y-plane
PostPosted: Wed Nov 16, 2011 11:11 am 
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questioner wrote:
The question does not mention anything about x & y being positive only.
Yes, it does.
Quote:
... , what is the probability that x > y > 0?
Sometimes some little information in a question statement slips the mind and it can affect the whole reasoning and lead to a wrong answer choice. Reading every word and symbol in a question statement is very important.


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