Suppose you throw a dart at a circular target of radius 10 inches. Assuming that you hit the target and that the coordinates of the outcomes are chosen at random, find the probability that the dart falls (a) within 2 inches of the center. (b) within 2 inches of the rim. (c) within the first quadrant of the target. (d) within the first quadrant and within 2 inches of the rim.

a) The probability is 0.04

b) The probability is 0.36

c) The pprobability is 0,25

d) The probability is 0.09

Step-by-step explanation:

Lets calculate areas:

the target has a radius of 10 inces, hence the target area has a area on 10²*π = 100π square inches.

a) A circle of 2 inches of radius has an area of 2²π = 4π square inches, hence the probability of hitting that area is 4π/100π = 1/25 = 0.04

b) If the dart s within 2 inches of the rim, then it is not at distance 8 inches from the center (that is the complementary event). The probability for the dart to be at 8 inches of the center is 8²π/100π = 64/100 = 16/25 = 0.64, thus, the probability that the dart is at distance 2 or less from the rim is 1-0.64 = 0.36.

c) The first quadrant has an area exactly 4 times smaller than the area of the target (each quadrant has equal area), thus the probability for the dart to fall there is 1/4 = 0.25

d) If the dart is within 2 inches from the rim (which has probability 0.36 as we previously computed), then it will be equally likely for the dart to be in either of the 4 quadrants (the area that is within 2 inches from the rim forms a ring and it has equal area restricted on each quadrant). Therefore, the probability for the dice to be in the first qudrant and within 2 inches from the rim is 0.36*1/4 = 0.09.