1) The random circle can only intersect the circle of radius 5 if the centre of the random circle is at most 1 unit away from the circumference of the circle of radius 5. So basically we can draw two boundaries above and below the circle of radius 5. These boundaries are circles of radius = 6 (outermost boundary) and radius = 4(innermost boundary) which are centred at the origin (just like the circle of radius 5). So if a random circle has its centre in that region, it will intersect the circle of radius = 5. Now, how many random circles can we fit inside that region? How many random circles can be placed inside the circle of radius 10? Well, we can use areas to solve this problem. The total area is pi*10^2 = 100pi. The area formed in the boundary is pi*6^2 - pi*4^2 = 36pi - 16 pi = 20 pi. So, the probability of intersection = 20pi/100pi = 1/5 = 0.2.
2.) Can you think of the general formula that we used for part 1 if we didn't know the exact value of the radius R? It is (pi*(R+1)^2 - pi*(R-1)^2)/(pi*100). Take the expected value of this equation to obtain the final answer. You will find that the expected value of this works out to be 0.14.
Terminology:
A circle of radius r is a curve that consists of all points at distance r from the center of the circle.
A disk of radius r is the set of all points whose distance from its center is less than or equal to r .
Thus, a circle is the boundary of a disk.
There are two circles of radius 10 and 5. A random circle of radius 1, whose center is inside the larger circle, may or may not intersect the circle of radius 5.
1) We generate a random circle of radius 1, whose center is uniformly distributed inside a disk of radius 10 centered at the origin;
Find the probability that the random circle intersects a circle of radius r=5 , which is centered at the origin. (Give a numerical answer.)
2)Answer the same question as in Part (1) but for the case where r, instead of being 5, is the realized value of a random variable R that is uniformly distributed between 2 and 5. (Give a numerical answer.)
4 answers
Is the part 2 as 0.12?
@Anonymous, E[((R+1)^1 - (R-1)^2)/100] = E[4R/100] =1/25 * E[R] = 1/25 * 3.5 = 0.14.
Sorry, typo it should be (R+1)^2 there, not (R+1)^1
1 answer