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Bunuel please help with a detailed explanation.

Regards
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I am a bit confused why you would need a statement to figure this out at all.

If the circle is in the square, then we have the area of the square and the triangle.

Circle = x2
Triangle = π(1/2)x2

Let's say x=2 Ac=4 At=3.14

Probability =.86/2

Does that seam right?
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RMD007
A Circle, with its radius an integer, is inscribed in a square. What is the probability that a point randomly chosen inside the square will lie outside the circle?

1. Side of the square is a prime number.

2. Another square inscribed in the circle has side \(\sqrt{2}\)

You don't need any statements to answer the question. The area of the circle is π r^2, and the area of the square is (2r)^2 = 4r^2, since the diameter of the circle is equal in length to a side of the square, if the circle is inscribed in the square. To find the probability a point chosen randomly from within the square is inside the circle, we just work out what fraction of square the circle takes up - we divide the area of the circle by the area of the square. So a random point in the square is also in the circle with probability π*r^2 / 4*r^2 = π/4, and the probability that point is not in the circle is thus 1 - (π/4). The length of the radius doesn't matter.

edit - had included another comment, but misread statement 2 :)
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RMD007
A Circle, with its radius an integer, is inscribed in a square. What is the probability that a point randomly chosen inside the square will lie outside the circle?

1. Side of the square is a prime number.

2. Another square inscribed in the circle has side \(\sqrt{2}\)

Source : Self-made. :)

If a circle is inscribed in a square, is it mandatory for the circle to touch the sides of the square? Shouldn't this be mentioned?

This way, the answer should be E.
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RMD007
A Circle, with its radius an integer, is inscribed in a square. What is the probability that a point randomly chosen inside the square will lie outside the circle?

1. Side of the square is a prime number.

2. Another square inscribed in the circle has side \(\sqrt{2}\)

Source : Self-made. :)

The original problem statement is good enough to answer this question. We do not need the 2 statements at all, because in such a symmetric arrangement all the peripherial lengths will be proportionate to each other.

The required probability = 1 - Probability that the point will lie both inside the square and the circle = 1 - pi*r^2/(2r)^2= 1 - pi/4 = 0.215
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1) Side of square is prime
Clearly side = a ( prime ) which is = 2r as circle is inscribed

P( reqd) = {a^2 - (πa^2)/4 }/a^2
=> (4-π)/4
Definite Ans A/D

2) A square is again inscribed in circle of side √2
=> radius =√(2+2) = 2
and side =4
again P reqd = ( 4x4- π4 )/16
p(reqd) = (4-π)/4
So D is the Answer

Bunuel
HOWEVER QUES SHOULD CLEARLY SPECIFY THAT WE HAVE TO TAKE A POINT INSIDE OF FIRST SQUARE AND OUTSIDE OF CIRCLE.
THIS COULD BE MISUNDERSTOOD AS A POINT MAY BE TAKEN INSIDE OF THE INNER SQUARE THEN ANS WOULD HAVE BEEN DIFFERENT.




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