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

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by BTGmoderatorLU » Sun Jul 08, 2018 5:37 pm

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Answers

A

B

C

D

E

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Difficulty—

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

The OA is A.

I'm confused by this PS question. Can someone assist with it, please? Thanks in advance!
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by [email protected] » Sun Jul 08, 2018 6:25 pm
Hi All,

This question is fairly 'high-concept', but you can solve it with a drawing and a bit of logic.

When you graph the equation X^2 + Y^2 = 1, you will have a circle with a radius of 1 that is centered around the Origin. There are only four points on that circle that are integer values: (0,1), (1, 0), (0, -1) and (-1, 0). All of the other points are positive/negative fractional values. Knowing that, the ONLY way for the inequality Y > X + 1 to occur is when X is NEGATIVE and Y is POSITIVE. That outcome only occurs in the 2nd quadrant of the graph (which is 1/4 of the circle). While you might be unsure about whether every point in that quadrant would 'fit' the inequality or not, the answer choices ARE numbers - and the smallest of them is 1/4. Since there's no smaller possibility, 1/4 must be the answer.

Final Answer: A

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