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Manager  Joined: 26 Mar 2007
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The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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Question Stats: 58% (01:29) correct 42% (01:32) wrong based on 375 sessions

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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

Originally posted by kannn on 05 May 2011, 23:25.
Last edited by Bunuel on 23 Jul 2015, 02:50, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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9
3
Since x > y > 0. Means that x,y>0 first quadrant is the region concentrated.- probability here is 1/4.

In the first quadrant in half of the cases x>y and other half y>x. As you can draw a line y = x through the region,splitting it into halves. Thus 1/2 is the probability.

total = 1/2 * 1/4 = 1/8
A.
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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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Alternatively, the step by step manual approach is:

Consider all possible outcomes about sign of x,y and their positions

SIGN POSITION
x y
+ + x sup to y
+ - x inf to y
- + x equal to y
- -
Go through each of the 4 cases and determine the number of possible outcomes: you will get 3 + 1 +1 + 3 = 8

The number of desirable outcomes is 1 ( + + and x sup to y).

Hence probability is 1/8.

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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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Probability of both x and y to be positive is 1/4 (quadrant 1). Probability of x>y is 1/2. So the probability of x>y>0 is (1/4)*(1/2) is 1/8.
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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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3
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

The favorable region will only be on the first quadrant between lines
y=0 AND y=x

The angle formed between these two lines from the center(0,0) = 45 degrees. Thus, the favorable sector within the circle will be 45/360=1/8 of the entire circle.

Ans: "A"

*******************

A picture would have explained it better. I don't have the tool handy.
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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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I got it. OA is A.
One question, will the slope of line y=x matters? Like if a line is y=4x, probability will still be 1/8?
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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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I got it. OA is A.
One question, will the slope of line y=x matters? Like if a line is y=4x, probability will still be 1/8?

y=x is imperative.

If y=4x
For x=1, y=4; but we want x>y. Moreover, the sector formed by x-axis(y=0) AND y=4x won't be (1/8)th of the entire circle; it will be more than 1/8 as the angle formed by the x-axis and y=4x is greater than 45 degrees.
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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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So i guess for condition x>y>0 only line can be y=x with slope 1. Other lines of form y=mx where m is not equal to 1 will not satisfy the condition.
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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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1
x>y>0

x>0,y>0 can only be possible in first quadrant.

x>y>0 can only be possible in one half of the first quadrant
(imagine line x=y line passing through origin , which cuts the first quadrant into two equal parts.)

=> likely outcome = (1/2)((2*pi*r)/4)= (2*pi*r)/8

probability = likely outcomes/ total outcomes

= (2*pi*r/8)/(2*pi*r)

= 1/8

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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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1
The probability of x & y > 0 in a circle is 1/4 ( Out of 4 quadrants only 1st satisfies the condition)
Now since x>y , so there would be half such points in the first quadrant.
Hence prob = 1/2 * 1/4 = 1/8.

P.S This probability also includes half of points for which x=y. IMO this is an incomplete question as nothing is mentioned using which we can calculate the number of points having x=y.
P(x>y) + p (x=y) + p(x<y) = 1
But in this question, we are ignoring P(x=y).
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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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Point (A,B) is randomly selected inside of circle
x^2+y^2=1. What is the probability that A>B>0?

M sorry i don't have the options! Answer is 1/8 though but can somebody please explain how.
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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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1
nshang wrote:
Point (A,B) is randomly selected inside of circle
x^2+y^2=1. What is the probability that A>B>0?

M sorry i don't have the options! Answer is 1/8 though but can somebody please explain how.

X^2+Y^2= 1, means radius = 1 centre origin(0,0)

If both a and b > 0 , the point has to be in the Ist quadrant. That gives a probability of 1/4.

Out of the quarter circle in the first quadrant 1/2 the sector will have a> b and the other half will have a< b
therefore its 1/2*1/4 = 1/8
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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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Having a hard time speculating probable scenarios. Could someone please explain this with a picture for species like me? :D

Thanks.
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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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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

Having a hard time speculating probable scenarios. Could someone please explain this with a picture for species like me? :D

Thanks.

Check the image below: Equation of the blue line is y = x. x > y > 0 is the yellow region, which is 1/8 of the circle.

Check other Probability and Geometry questions.

Attachment: Untitled.png [ 7.66 KiB | Viewed 6704 times ]

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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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1
kannn wrote:
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

1st Quadrant, portion below x=y line is the region which satisfies x>y>0, which is 1/8 of the area of circle and required probability.

Refer image
Attachments

File comment: Solution by geometric division IMG_20171002_220352.jpg [ 1.21 MiB | Viewed 4066 times ]

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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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[quote="Bunuel"][quote="ranaazad"]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

If this statement was just X>Y and didn't include the 0 limitation, then would the answer be 1/2?
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Posts: 58428
Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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Bunuel wrote:
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

If this statement was just X>Y and didn't include the 0 limitation, then would the answer be 1/2?

Right. In this case the region would be under the blue line, which is 1/2 of the circle. _________________
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Re: The center of a circle lies on the origin of the coordinate plane. If  [#permalink]

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_________________ Re: The center of a circle lies on the origin of the coordinate plane. If   [#permalink] 25 Apr 2019, 08:03
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