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# A square in the XY co-ordinate system has vertices as (1,1),

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CEO
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A square in the XY co-ordinate system has vertices as (1,1), [#permalink]  15 Sep 2003, 07:01
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A square in the XY co-ordinate system has vertices as (1,1), (1,-1), (-1,-1), (-1,1). If a is the x coordinate of any point and b the y-coordinate of any point in the XY coordinate system, find the probability that a point selected at random from the square region will satisfy x^2+ y^2<1?

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Praetorian
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Re: PS : probability [#permalink]  15 Sep 2003, 07:22
praetorian123 wrote:
A square in the XY co-ordinate system has vertices as (1,1), (1,-1), (-1,-1), (-1,1). If a is the x coordinate of any point and b the y-coordinate of any point in the XY coordinate system, find the probability that a point selected at random from the square region will satisfy x^2+ y^2<1?

Thanks
Praetorian

Not sure if this is correct ..

x^2 + y^2 =1 is a circle around (0,0) with a radius of 1. And any point within that circle should have x^2 + y^2 < 1. So, probability of a point selected randomly that will satisfy x^2 + y^2 < 1 is:

Area of Circle/Area of Square = pi/4 = 3.14/4 = 0.785
CEO
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Re: PS : probability [#permalink]  15 Sep 2003, 07:35
edealfan wrote:

Not sure if this is correct ..

x^2 + y^2 =1 is a circle around (0,0) with a radius of 1. And any point within that circle should have x^2 + y^2 < 1. So, probability of a point selected randomly that will satisfy x^2 + y^2 < 1 is:

Area of Circle/Area of Square = pi/4 = 3.14/4 = 0.785

See the attachment

Area of Square 1 / Area of Square 2 = [sqrt(2)]^2 /(2^2) = 2/4 =0.5

But here we assumed x^2 + y^2 =1

So any value lesser than 0.5 should be a candidate..

Intern
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where does the circle come into the picture ? There is no need to get into circles for a simple question like this.

The questions asks of 4 points of the square. Hence the P=0.
Manager
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I see the trick here...

You basically have to draw lines x = 1 - y, x = y - 1 and the same set for y in terms of x. Doing this will give you a square between the four lines. You have to recognize that in order for the original equation (x^2 + y^2 < 1) to be true, you have to consider the points inside the new square.

Now targeting the probability question, you can see that the area of the new square is half that of the original square. So of ALL the points in the bigger square, half has many will fall in the new square. This will give you probability of 1/2. Well, technically the probability is a little less than 1/2 but it's so close that we can just say 1/2.

Good question.
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wonder_gmat wrote:
I see the trick here...

You basically have to draw lines x = 1 - y, x = y - 1 and the same set for y in terms of x. Doing this will give you a square between the four lines. You have to recognize that in order for the original equation (x^2 + y^2 < 1) to be true, you have to consider the points inside the new square.

Now targeting the probability question, you can see that the area of the new square is half that of the original square. So of ALL the points in the bigger square, half has many will fall in the new square. This will give you probability of 1/2. Well, technically the probability is a little less than 1/2 but it's so close that we can just say 1/2.

Good question.

Correct me if I am wrong, but here is a problem with your solution.

With the new square you created within the given square, consider a point say (1/2, 1/2). It is on the line x = 1 - y or x + y = 1. Now, consider a point, (0.55, 0.55) which is a little above (1/2, 1/2).

Now, (0.55)^2 + (0.55)^2 = 0.605 (which is less than 1).

x^2 + y^2 = 1 is a circle and not a square. Hope this helps.
CEO
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edealfan wrote:
Correct me if I am wrong, but here is a problem with your solution.

With the new square you created within the given square, consider a point say (1/2, 1/2). It is on the line x = 1 - y or x + y = 1. Now, consider a point, (0.55, 0.55) which is a little above (1/2, 1/2).

Now, (0.55)^2 + (0.55)^2 = 0.605 (which is less than 1).

x^2 + y^2 = 1 is a circle and not a square. Hope this helps.

Watchdog:
If you explain your solution, maybe we can tell you if you right.

Wonder :
I constructed the square using the pythagorous theorem for each set of points. for eg, for (1,1) ....1^2 + 1^2 = 2
So the Hypotenuse or the side of the square = Sqrt(2)

Edeal :
GMAT never tests on Circle formulas...so i was trying to see if anyone can do in a simpler way..
But i guess we can leave it at that...we have to rely on the answer choices for a definite answer.

Thanks
Praetorian
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# A square in the XY co-ordinate system has vertices as (1,1),

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