Probability that circle of a certain radius lies within a rectangle. A $10\times 10$ square card is folded to create two $5\times 10$ rectangles. The upper half of the square is blue and the lower half of the square is red. What is the probability that a circle of radius $1$ lies entirely in the red part?
 A: We solve the problem on the assumption that the circle's centre is uniformly distributed in the $8\times 8$ square drawn by forbidding all locations that would put parts of the circle outside the square.
To get the circle entirely in the red rectangle, we need to confine the centre to a $3\times $8$ rectangle. 
The required probability is therefore $\frac{3\times 8}{8\times 8}$.
A: Assuming two things:


*

*The position of the centre of the circle is randomly chosen, i.e. uniformly distributed on the $10\times 10$ square:

*For the circle of radius 1 to lie entirely in the red, its edges may lie tangent to the edges of the card.


Then it's not too hard! On a $5\times 10$ rectangle, this means that there's a smaller rectangle within, which is the set of eligible locations for the center of the circle. This smaller rectangle is chosen such that if the center of a circle were right on its borders, the edge of the circle would touch the edge of the paper. Here's a diagram. The eligible area is the green rectangle:

So, if we add a border of 1 to allow for the radius of 1 on all sides of the rectangle, then we get a rectangle that measures $3\times 8$. That's a total area of $24$ units$^{2}$, and your original square had an area of $10\times 10 = 100$ units$^{2}$. $24/100$ gives us a probability of $0.24$.
Note that my answer disagrees with that of Andre Nicolas, because our assumptions are different. You need to double-check whether the circle is allowed to go outside the square, or is positioned totally inside the paper square.
