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PriyankaPalit7
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Point (A,B) is randomly selected inside of circle x^2+y^2=1. What is t
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14 Mar 2019, 09:38
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65% (hard)

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44% (01:24) correct 56% (01:16) wrong based on 201 sessions

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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?

A) \(1/4\)
B) \(1/8\)
C) \(1/6\)
D) \(1/2\)
E) \(1\)
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iPrasad
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Point (A,B) is randomly selected inside of circle x^2+y^2=1. What is t
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19 Mar 2019, 12:09
As per the attached picture answer should be 1/8
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Point (A,B) is randomly selected inside of circle x^2+y^2=1. What is t
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19 Mar 2019, 10:30
PriyankaPalit7
Point (A,B) is randomly selected inside of circle \(x^2+y^2\)=1. What is the probability that A>B>0?

A) \(1/4\)
B) \(1/8\)
C) \(1/6\)
D) \(1/2\)
E) \(1\)
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area of circle = pi * r^2
r=1 so area of pi
so only at 1st quadrant a>b>0 would be valid
1/4 ; should be answer
PriyankaPalit7
is given answer option correct?
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PriyankaPalit7
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Joined: 28 May 2018
Last visit: 13 Jan 2020
Posts: 116
Location: India
Schools: ISB '21 (A)
GMAT 1: 640 Q45 V35
GMAT 2: 670 Q45 V37
GMAT 3: 730 Q50 V40
Schools: ISB '21 (A)
GMAT 3: 730 Q50 V40
Posts: 116
Point (A,B) is randomly selected inside of circle x^2+y^2=1. What is t
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20 Mar 2019, 03:10
Archit3110
PriyankaPalit7
Point (A,B) is randomly selected inside of circle \(x^2+y^2\)=1. What is the probability that A>B>0?

A) \(1/4\)
B) \(1/8\)
C) \(1/6\)
D) \(1/2\)
E) \(1\)

area of circle = pi * r^2
r=1 so area of pi
so only at 1st quadrant a>b>0 would be valid
1/4 ; should be answer
PriyankaPalit7
is given answer option correct?
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Quadrant i would have A and B > 0. ie. positive values for A and B.
But only half of quadrant 1 would have A > B. SO the correct answer is B 1/8.
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anubhv
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Point (A,B) is randomly selected inside of circle x^2+y^2=1. What is t
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16 Sep 2019, 09:02
PriyankaPalit7
Point (A,B) is randomly selected inside of circle \(x^2+y^2\)=1. What is the probability that A>B>0?

A) \(1/4\)
B) \(1/8\)
C) \(1/6\)
D) \(1/2\)
E) \(1\)
Show more

You are also including the probability of the point lying on y=x and y=0 as you are considering those points also in the area, which should not be the case as a cannot be equal to b and y cannot be equal to 0
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Tanisha2819
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Point (A,B) is randomly selected inside of circle x^2+y^2=1. What is t
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13 May 2023, 13:11
Bunuel KarishmaB @VeritasPrep

Could any of the experts please help me by explaining the solution of this question? I am unable to solve this.

Thank you
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KarishmaB
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Point (A,B) is randomly selected inside of circle x^2+y^2=1. What is t
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13 May 2023, 19:28
Tanisha2819
Bunuel KarishmaB @VeritasPrep

Could any of the experts please help me by explaining the solution of this question? I am unable to solve this.

Thank you
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The circle x^2 + y^2 = 1 lies with its centre at (0, 0) and has radius 1. So it is as shown below.
Attachment:
Screenshot 2023-05-14 at 8.56.55 AM.png
Screenshot 2023-05-14 at 8.56.55 AM.png [ 40.71 KiB | Viewed 2865 times ]

Since we want X > Y > 0, we are looking for points for which both x and y co-ordinates are positive. Those lie in the first quadrant.
The line y = x where x co-ordinate is equal to y co-ordinate divides the XY plane into two parts. On its right side, x > y (green region) and on its left side x < y (orange region).

So half of the quarter region inside the circle in the first quadrant (darkened) represents the area in which for all points, X > Y. Hence answer is 1/8.

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Point (A,B) is randomly selected inside of circle x^2+y^2=1. What is t
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