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probability that b>a+1

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by kaulnikhil » Thu May 20, 2010 7:27 am
Set T consists of all points (x,y) such that x^2+y^2=1 . If point (a,b) is selected from set T at random, what is the probability that b>a+1?

1/4
1/2
1/3
3/5
2/3
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by gmatmachoman » Thu May 20, 2010 7:36 am
(x,y) can be

(1,0)

(0,1)

(-1,0)

(0,-1)

For b>a+1

(-1,1) is the only set that satisfie sthe condition.

p = 1/4.
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by liferocks » Thu May 20, 2010 7:41 am
x^2+y^2=1 is the equation of circle with radius 1 and center 0...
since a,b is on the circle -1<a<1 and -1<b<1
so b>a+1 or b-a>1 is only possible when the point is in second quadrant

Since the circle x^2+y^2=1 is equally divided in to 4 parts by the X and Y axis the probability that point (a,b) is in second quadrant is [spoiler]1/4[/spoiler]
Ans option A
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by kaulnikhil » Thu May 20, 2010 8:47 am
liferocks wrote:x^2+y^2=1 is the equation of circle with radius 1 and center 0...
since a,b is on the circle -1<a<1 and -1<b<1
so b>a+1 or b-a>1 is only possible when the point is in second quadrant

Since the circle x^2+y^2=1 is equally divided in to 4 parts by the X and Y axis the probability that point (a,b) is in second quadrant is [spoiler]1/4[/spoiler]
Ans option A
I got a doubt ....
Let us consider the equation as b=a+1
now this equations cuts the circle at (-1,0 and 0,1).. Now in case B>a+1
arnt we supposed to find the area of the region enclosed the second quadrant by the line and the circle ?? the area i am referring is above the line b=a+1 ans below the circular region x^2+y^2=1 in second quadrant.. In that case the probablity has to be less than 1/4
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by liferocks » Thu May 20, 2010 9:12 am
kaulnikhil wrote:
liferocks wrote:x^2+y^2=1 is the equation of circle with radius 1 and center 0...
since a,b is on the circle -1<a<1 and -1<b<1
so b>a+1 or b-a>1 is only possible when the point is in second quadrant

Since the circle x^2+y^2=1 is equally divided in to 4 parts by the X and Y axis the probability that point (a,b) is in second quadrant is [spoiler]1/4[/spoiler]
Ans option A
I got a doubt ....
Let us consider the equation as b=a+1
now this equations cuts the circle at (-1,0 and 0,1).. Now in case B>a+1
arnt we supposed to find the area of the region enclosed the second quadrant by the line and the circle ?? the area i am referring is above the line b=a+1 ans below the circular region x^2+y^2=1 in second quadrant.. In that case the probablity has to be less than 1/4
x^2+y^2=1 is the points on the circle not inside the circle so we don't need the area. You are correct for the point (-1,0) b-a=1 not b-a>1..but second quadrant has infinite number of points so one point not going to make any difference probalility will be 1/4 or almost 1/4.
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by frank1 » Mon Jun 07, 2010 7:15 pm
gmatmachoman wrote:(x,y) can be

(1,0)

(0,1)

(-1,0)

(0,-1)

For b>a+1

(-1,1) is the only set that satisfie sthe condition.

p = 1/4.
could you please clarify?
from where that -1,1 came...
it doesnt safisfy the condition as well
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by jahnaviyeruva » Tue Jun 08, 2010 2:30 am
i guess the answer is 1/3
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by nmerchant » Tue Jun 08, 2010 11:57 am
The correct answer is p=1/4

Here's how. The only way b > a+1 is if a is negative and b is positive.
This means all the points in the second quadrant that satisfy a^2 + b^2 = 1 will satisfy b > a + 1.
For example lets substitute a = -0.8 and b = 0.6
0.6 > -0.8 + 1 and -0.8^2 + 0.6^2 = 1

So the solution set is 1/4th of the total number if items in Set T. Hence the probability is 1/4
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