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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
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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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05 May 2011, 23:54
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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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06 May 2011, 01:34
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. Answer is 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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16 Sep 2011, 05:35
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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16 Sep 2011, 05:36
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jamifahad 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 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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16 Sep 2011, 10:39
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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16 Sep 2011, 11:16
jamifahad wrote: 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 xaxis(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 xaxis 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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16 Sep 2011, 14:08
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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16 Sep 2011, 19:58
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
Answer is 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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17 Sep 2011, 23:24
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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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04 Oct 2011, 23:34
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. Thanks in advance!



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Re: The center of a circle lies on the origin of the coordinate plane. If [#permalink]
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04 Oct 2011, 23:57
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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. Thanks in advance! 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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13 Sep 2016, 08:16
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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Re: The center of a circle lies on the origin of the coordinate plane. If [#permalink]
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02 Oct 2017, 09:39
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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
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Re: The center of a circle lies on the origin of the coordinate plane. If [#permalink]
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07 Apr 2018, 12:40
[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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Re: The center of a circle lies on the origin of the coordinate plane. If [#permalink]
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