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Geometric Probability 11.5. Objectives: Solve problems involving geometric probability. Solve problems involving sectors and segments of circles.

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Presentation on theme: "Geometric Probability 11.5. Objectives: Solve problems involving geometric probability. Solve problems involving sectors and segments of circles."— Presentation transcript:

1 Geometric Probability 11.5

2 Objectives: Solve problems involving geometric probability. Solve problems involving sectors and segments of circles.

3 [ Probability is expressed as a fraction, decimal, or percent.] Geometric Probability- Probability that involves a geometric measure such as length or area. If a point in Region A is chosen at random, then the probability, P(B), that the point is in region B which is in the interior of Region A is P(B)= area of region B area of region A A B

4 Example 1: What is the chance that a dart will land in the yellow area of the grid? Find the area of the whole grid. To find the area of the yellow area, count the square units. Yellow area: 28 Grid area: 49 P(yellow)= Which would be: 4/7 or ≈.6

5 Example 1 continued: What is the chance that a dart will land in the yellow area of the grid? Find the area of the whole grid. (7 times 7) To find the area of the yellow area, count the square units. Yellow area: 28 Grid area: 49 P(yellow)= Which would be: 4/7 or ≈.6

6 Your Turn: Find the probability that a point chosen at random will land in the green area. Region A: Region B:

7 How to solve: First, we have to find the area of region A. Then, we find the area of region B. P(B)= 9 81 So P(B)= 1 9 So therefore, the probability that a point will land in region B is 1/9 or ≈.11

8 - A sector of a circle is a region of a circle bounded by a central angle and its intercepted arc. Sector Intercepted Arc Central Angle

9 Area of a Sector If a sector of a circle has an area of A square units, a central angle measuring N degrees, and a radius of r units, then Sector N

10 Probability with Sectors We know how to calculate the area of a sector but we still have to find the probability that a point chosen at random will land in that sector. The way you do this is by using the following formula: P = area of sector area of circle

11 Example 2: Find the area of sector XYZ and then find the probability that a point chosen at random will land in sector XYZ. X Y Z

12 How To: Put the central angle over 360 and multiply it to pi and r². A= 120/360(π)(5²) A= 120/360(3.14)(25) A= 26.2cm² To find the probability: Area of a circle is πr². So, π(5²) = π(25) *calculator* Area of circle is ≈78.5cm² P = Probability that a random point is in sector XYZ is about.33 or 33%

13 Find the measure of X, the area of the sector ABC, and find the probability that a point chosen at random will land in sector ABC. X degrees 123° 16m B A C X = 57 A =31.8m² P ≈.16 X degrees= 57° Area of sector: 57 (3.14)(8²) 360 A≈ 31.8m² Area of circle: (3.14)(8²) A≈ P =

14 The region of a circle bounded by an arc and a chord is called a segment of a circle. Chord Segment Arc

15 Area/Probability of a Segment In any regular polygon inscribed in a circle, to find the area of a segment, you first have to: -Find the area of a sector. -Find the area of a triangle. Then, you can find the area of the segment. -Area of Segment= area of sector- area of triangle To find the probability that a point chosen at random lies in the segment is: P = area of segment area of circle

16 Example 3: Find the area of one of the blue segments in the circle and the probability that a point chosen at random will land in the segment. 6cm Area of sector: 60 (π)(3²) 360 ≈ 4.7 cm² Area of triangle: (1/2)(bh) (1/2)(3)(h) The are equilateral triangles, so the apothem would make a 30°-60°-90° triangle. Therefore, the height would be 1.5√3. A = (1/2)(3)(1.5√3) = ≈ 3.9cm² A of segment= 4.7 – 3.9 A = 0.8cm² Area of circle: (3.14)(3²) ≈ 28.3 P = ≈.03 or 3% 60° 30° 1.5cm 1.5√3

17 30 in. Area of Sector: 120 πr² 360 A = 120/360 (3.14)(15²) = 235.5in² Area of circle: 15²(3.14)= 706.5in² 15in. 30° 60° Each central angle is 120° Apothem= 7.5 Base of triangle= 7.5√3 ≈13in 7.5 in. 13 in. Area of triangle: ½(26)(7.5) =97.5in² Area of Segment: – 97.5 = 138in² P = ≈.2 or 20% Find the area of the shaded region. Then, find the probability that a point chosen at random will land in the green area.

18 ASSIGNMENT!!!!! Page : 7-18, 20-25, 34


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