Question

Let O(0,0),A(6,0),B(6,6),C(0,6) be the vertices of a square OABC, and let M be the midpoint of OB. Find the probability that a point chosen at random from the square is: a farther from O than from M, b more than twice as far from O as from M.

Answer 1

The probability that a point chosen at random from the square is farther from O than from M is 0.125.

Step-by-step explanation:

To find the probability that a point chosen at random from the square is farther from O than from M, we need to find the area of the region that satisfies this condition. The square OABC has side length 6 units, so its area is 6*6 = 36 square units.

The line segment OM has length 6/2 = 3 units, which is the distance from O to the midpoint of OB. The region where a point is farther from O than from M is the upper right triangle OMB.

The area of the triangle OMB is (1/2)*(base)*(height) = (1/2)*(3)*(3) = 4.5 square units.

Therefore, the probability that a point chosen at random from the square is farther from O than from M is the ratio of the area of the triangle OMB to the area of the square OABC, which is 4.5/36 = 1/8 = 0.125.