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EXAMPLE 3

ARCHERY. EXAMPLE 3. Use areas to find a geometric probability.

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EXAMPLE 3

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  1. ARCHERY 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?

  2. Area of red circle P(arrow lands in red region) = Area of target (82) 64 1 = = = (402) 1600 25 ANSWER The probability that the arrow lands in the red region is 1 , or 4%. 25 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.

  3. SCALE DRAWING Find: the area of the field. The shape is a rectangle, so the area is bh = 10 3 = 30 square units. EXAMPLE 4 Estimate area on a grid to find a probability 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

  4. Find: the total area of the park. EXAMPLE 4 Estimate area on a grid to find a probability STEP 2 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. 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 52 + 6.5 =58.5 square units for the total area.

  5. Area of field 300 30 20 P(ball in field) = = = Total area of park 58.5 39 585 The probability that the ball is in the field is about 20 , or 51.3%. ANSWER 39 EXAMPLE 4 Estimate area on a grid to find a probability STEP 3 Write: a ratio of the areas to find the probability.

  6. A = πr2 A = π(8)2 A = 64π square centimeters for Examples 3 and 4 GUIDED PRACTICE 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

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

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

  9. A = π(8+8+8+8+8)2 – π(8+8+8+8)2 A = 1600π – 1024π A = 576π square centimeters A = 576π + 320π A = 896πsquare centimeters for Examples 3 and 4 GUIDED PRACTICE STEP 5 Repeat the process for the outer black ring. STEP 6 To compute total area of the black rings, add.

  10. 896π 14 = 1600π 25 for Examples 3 and 4 GUIDED PRACTICE STEP 7 Compute the ratio – the total area is π(5·8)2. (same as the area through the outer black ring). = 56%

  11. – Area of woods = Area of park Area of field – = 58.5 30 Area of woods 28.5 19 P(ball in woods) = = = Total area of park 58.5 39 for Examples 3 and 4 GUIDED PRACTICE 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. STEP 2 Write a ratio of the areas to find the probability. or 48.7%.

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