Question

Let (X,Y) denote the coordinates of a point selected at random inside the unit circle whose center is at the origin. That is, X and Y have a joint density function given by

f (x.y)= {1/π when x² + y² ≤ 1
{ 0 otherwise

Find P[Y > X).

Answer 1

To find P[Y > X], we need to calculate the probability that Y is greater than X. Using the given joint density function, we can integrate over the region where Y is less than X in the bottom half of the unit circle to find this probability.

Step-by-step explanation:

To find P[Y > X], we need to calculate the probability that Y is greater than X. In this case, Y represents the vertical coordinate and X represents the horizontal coordinate of a point selected randomly inside the unit circle. The joint density function given is f(x,y) = 1/π when x² + y² ≤ 1, and 0 otherwise.

To calculate P[Y > X], we need to determine the region where Y is greater than X. In the unit circle, Y will be greater than X in the top half of the circle.

Since the unit circle has symmetry, we can determine the probability of Y being greater than X in the top half of the circle by calculating the probability of Y being less than X in the bottom half of the circle. Therefore, P[Y > X] = 1 - P[Y < X].

Using the symmetry of the unit circle, we can see that the region where Y is less than X in the bottom half of the circle is the same as the region where X is less than Y in the top half of the circle. Therefore, P[Y < X] = P[X < Y].

Using the given joint density function, we can calculate this probability by integrating the joint density function over the region where X is less than Y in the top half of the circle.

Since the density function is defined as 1/π when x² + y² ≤ 1, and 0 otherwise, we integrate over the region where x² + y² ≤ 1 and y > x in the top half of the circle.