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Assume the sides of the rectangle to be 2 and 4.
Diameter of the circle=AB=2
Radius of the circle=1
Area of circle= {\pi}
Area of the rectangle = 2*4=8
Area of the rectangle outside circle = 8-{\pi}
So, probability= 8-{\pi}/8
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Let us assume the sides of the rectangle be 1 and 2, so the area of the rectangle is 2 which implies that the circle is inscribed within a square whose area is 1.

Area of circle inscribed within a square is \(\frac{pi}{4}\) times the area of square = \(\frac{pi}{4}\)

Probability of point not inside the circle = 1 - probability of point inside the circle = 1 - \((pi/4)/2\) = 1 - \(\frac{pi}{8}\) = \(\frac{(8-pi)}{8}\)


p.s how does one type the symbol of pi?
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nave81
p.s how does one type the symbol of pi?
Dear nave81 ---

It's funny. I was wondering this same thing, and I had to quote a response of Bunuel in which he used the \({\pi}\) symbol to see what it looked like in the html text.

Basically, you type {\pi} ----- (open curvy brackets)(backstroke)("pi")(close curvy brackets) ---- and then highlight that in the "math" delimiters ---- the m button, under the bold button in the rtf bar at the top of the editing window, does this. All math symbols need to be within the "math" delimiters.

Does this make sense?

Mike :-)
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Good question kudos given!

Probability = (Area of rectangle - Area of circle) / area of rectangle

b = breadth , l = length, r = radius.
we know l = 2b so area of rectangle = l * b = 2b^2
.
we can see the diameter of the circle = b hence radius = b/2
area will be π(b^2) / 4

Simplifying we get (8 - π)/8
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Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

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