EXAMPLE 3 Use areas to find a geometric probability The diameter of the target shown at the right is 80 centimeters. The diameter of the red circle on the target is 16 centimeters. An arrow is shot and hits the target. If the arrow is equally likely to land on any point on the target, what is the probability that it lands in the red circle? ARCHERY

EXAMPLE 3 Use areas to find a geometric probability SOLUTION Find the ratio of the area of the red circle to the area of the target. P (arrow lands in red region) = Area of red circle Area of target = (8 2 ) (40 2 ) = = 1 25 The probability that the arrow lands in the red region is 1 25, or 4%. ANSWER

EXAMPLE 4 Estimate area on a grid to find a probability SCALE DRAWING Your dog dropped a ball in a park. A scale drawing of the park is shown. If the ball is equally likely to be anywhere in the park, estimate the probability that it is in the field. SOLUTION STEP 1 Find: the area of the field. The shape is a rectangle, so the area is bh = 10 3 = 30 square units.

EXAMPLE 4 STEP 2 Find: the total area of the park. Make groups of partially covered squares so the combined area of each group is about 1 square unit. The total area of the partial squares is about 6 or 7 square units. So, use = 58.5 square units for the total area. Estimate area on a grid to find a probability Count: the squares that are fully covered.There are 30 squares in the field and 22 in the woods. So, there are 52 full squares.

EXAMPLE 4 STEP 3 Write: a ratio of the areas to find the probability. Estimate area on a grid to find a probability P (ball in field) = Area of field Total area of park = = The probability that the ball is in the field is about 20 39, or 51.3%. ANSWER

GUIDED PRACTICE for Examples 3 and 4 6. In the target in Example 3, each ring is 8 centimeters wide. Find the probability that an arrow lands in the black region. SOLUTION STEP 1 Compute the area of the red circle A = πr 2 A = π(8) 2 A = 64π square centimeters

GUIDED PRACTICE for Examples 3 and 4 STEP 2 Each ring adds 8 centimeters to the radius of the red circle. The area of the red circle + the white ring around it would be A = π(8 + 8) 2 A through the first White Ring = π(8+8) 2 A = π(16) 2 A = 256π square centimeters

GUIDED PRACTICE for Examples 3 and 4 STEP 4 To compute the area of just the black ring, take the area through the first black ring and subtract the area through the first White Ring. STEP 3 The area of the red circle + the white ring around it + the black ring would be A = π( ) 2. A through the first Black Ring = π(8+8+8) 2 A = π(24) 2 A = 576π square centimetersA of the first Black Ring = π(8+8+8) 2 – π(8+8) 2 A = 576π – 256πA = 320π square centimeters

GUIDED PRACTICE for Examples 3 and 4 STEP 6 To compute total area of the black rings, add. STEP 5 Repeat the process for the outer black ring. A = π( ) 2 – π( ) 2 A = 1600π – 1024πA = 576π square centimetersA = 576π + 320π A = 896π square centimeters

GUIDED PRACTICE for Examples 3 and 4 STEP 7 Compute the ratio – the total area is π(5·8) 2. (same as the area through the outer black ring). 896π 1600π = = 56%

GUIDED PRACTICE for Examples 3 and 4 7. In Example 4, estimate the probability that the ball is in the woods. SOLUTION STEP 1 Find the area of the woods. = 28.5 square units. = – Area of woods = Area of park Area of field – STEP 2 Write a ratio of the areas to find the probability. P (ball in woods) = Area of woods Total area of park = or 48.7%. = 19 39