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.)
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.
@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
Is the part 2 as 0.12?
To find the probability that the random circle intersects a circle of radius 5, we need to consider the possible positions of the center of the random circle within the disk of radius 10 centered at the origin.
1) Given that the random circle has a radius of 1 and its center is uniformly distributed within the disk of radius 10, we can calculate the area of the disk of radius 10 as π * (10^2) = 100π.
To determine the probability of intersection, we need to consider the range of positions where the center of the random circle can lie such that it intersects the circle of radius 5 centered at the origin.
If the center of the random circle lies within a disk of radius 6 centered at the origin, it will intersect the circle of radius 5.
The area of the disk of radius 6 is π * (6^2) = 36π.
Therefore, the probability that the random circle intersects the circle of radius 5 is equal to the ratio of the area of the disk of radius 6 to the area of the disk of radius 10:
Probability = Area of the disk of radius 6 / Area of the disk of radius 10
= (36π) / (100π)
= 0.36
So the probability that the random circle intersects the circle of radius 5 is 0.36.
2) In this case, instead of the circle of radius 5 being fixed, its radius r is a random variable R that is uniformly distributed between 2 and 5.
To find the probability in this scenario, we need to consider the range of possible values for R and calculate the probability for each value.
The possible values for R range from 2 to 5 (inclusive). For each value of R, we can calculate the probability of intersection as done in the first part.
Let's denote the probability that the random circle intersects the circle of radius R as P(R):
P(R) = Area of the disk of radius R + 1 / Area of the disk of radius 10
So, the probability of intersection for the case where R is uniformly distributed between 2 and 5 is given by integrating P(R) over the range [2, 5]:
Probability = ∫[2, 5] P(R) dR
To obtain a numerical answer, we need to evaluate this integral using appropriate techniques for numerical integration.
Please note that for this specific calculation, it is recommended to use mathematical software or programming to calculate the definite integral.