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Post subject: GMAT Coordinate Geometry Posted: Sat Jul 10, 2010 5:24 am 

Joined: Sun May 30, 2010 3:15 am Posts: 424

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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Gennadiy

Post subject: Re: math (test 1, question 26): probability, x,yplane Posted: Sat Jul 10, 2010 6:11 am 

Joined: Sun May 30, 2010 2:23 am Posts: 498

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:
Area equals πr²
Satisfactory events:
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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questioner

Post subject: Re: math (test 1, question 26): probability, x,yplane Posted: Sun Nov 28, 2010 5:26 pm 

Joined: Sun May 30, 2010 3:15 am Posts: 424

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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Gennadiy

Post subject: Re: math (test 1, question 26): probability, x,yplane Posted: Sun Nov 28, 2010 5:37 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498

When we deal with probability on the x,yplane 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: )


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Ashot

Post subject: Re: math (test 1, question 26): probability, x,yplane Posted: Sun Aug 28, 2011 3:47 pm 

Joined: Sun Aug 28, 2011 3:42 pm Posts: 1

How do we get 1/8 for the area of shaded region?
Thanks in Advance!


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Gennadiy

Post subject: Re: math (test 1, question 26): probability, x,yplane Posted: Sun Aug 28, 2011 4:39 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498

The xaxis and yaxis divide the circle into quarters (1/4).
The line y = x divides the 90⁰angle (between the coordinate axes) in half, into two 45⁰angles. 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.


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questioner

Post subject: Re: math (test 1, question 26): probability, x,yplane Posted: Tue Oct 11, 2011 11:33 am 

Joined: Sun May 30, 2010 3:15 am Posts: 424

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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Gennadiy

Post subject: Re: math (test 1, question 26): probability, x,yplane Posted: Tue Oct 11, 2011 11:36 am 

Joined: Sun May 30, 2010 2:23 am Posts: 498


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questioner

Post subject: Re: math (test 1, question 26): probability, x,yplane Posted: Wed Nov 16, 2011 11:01 am 

Joined: Sun May 30, 2010 3:15 am Posts: 424

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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Gennadiy

Post subject: Re: math (test 1, question 26): probability, x,yplane Posted: Wed Nov 16, 2011 11:11 am 

Joined: Sun May 30, 2010 2:23 am Posts: 498


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