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Set T consists of all points (x,y) such that x^2+y^2=1.

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by GMATGuruNY » Mon Apr 25, 2016 2:31 pm
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?

A.1/4
B.1/3
C.1/2
D.3/5
E.2/3
x² + y² = r² is the equation of a circle with its center at the origin and a radius of r.
Thus:
x² + y² = 1 is the equation of a circle with its center at the origin and a radius of 1.

Draw the circle and the line y = x+1:
Image

Every point in the yellow region is such that y > x+1.
Implication:
If point (a, b) on the circle is within the yellow region, then b > a+1.
Since 1/4 of the circle is within the yellow region, the probability that b > a+1 is 1/4.

The correct answer is A.
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by sachin_yadav » Tue Apr 26, 2016 9:30 am
Mitch,

Thanks for your reply.

But i am not able to understand 1/4. As you have mentioned "it is 1/4 of the circle within the yellow region".

However, it is part of 1/4 of the circle i.e x of 1/4 of circle

I am not clear about 1/4. I think you are talking about the sector.

Can you please clarify.

Sachin
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by GMATGuruNY » Tue Apr 26, 2016 9:42 am
sachin_yadav wrote:Mitch,

Thanks for your reply.

But i am not able to understand 1/4. As you have mentioned "it is 1/4 of the circle within the yellow region".

However, it is part of 1/4 of the circle i.e x of 1/4 of circle

I am not clear about 1/4. I think you are talking about the sector.

Can you please clarify.

Sachin
The yellow region includes all points such that y > x + 1.
x² + y² = 1 represents all points ON THE CIRCLE.
Of all points (a, b) on the circle such that a² + b² = 1, only those on the minor arc between (-1, 0) and (0, 1) are within the yellow region, implying that b > a+1.
Since this minor arc constitutes 1/4 of the circumference, the probability that b > a+1 is 1/4.
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by Matt@VeritasPrep » Fri Apr 29, 2016 3:03 pm
We could also do this algebraically.

Since a² + b² = 1, we must have 1 ≥ b ≥ -1 and 1 ≥ a ≥ -1. If we don't have both of these, then a² + b² > 1.

Since we seek b > a + 1, we also seek 0 > a ≥ -1. If a > 0, then b > 1, but we we learned above that 1 ≥ b, so we must have 0 > a.

That means our question is really "What's the probability that 1 ≥ b > 0 and 0 > a ≥ -1?"

Since we're drawing b from the set 1 ≥ b ≥ -1, there's a 1/2 chance that b is positive.

Since we're drawing a from the set 1 ≥ a ≥ -1, there's a 1/2 chance that a is negative.

(1/2) * (1/2) = 1/4, so there's a 1/4 chance that we get b > 0 and 0 > a.
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by sachin_yadav » Sun May 01, 2016 3:47 am
Matt,

When you say there is half chance that b is positive or b is negative. I hope you mean the below in the diagram and same for a.
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half chance b is positive.jpg
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by sachin_yadav » Thu May 05, 2016 10:00 am
Will greatly appreciate if anyone can tell me that i am correct in my thinking
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by Matt@VeritasPrep » Wed May 11, 2016 11:14 pm
sachin_yadav wrote:Will greatly appreciate if anyone can tell me that i am correct in my thinking
You are correct! :)
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