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11.5 Geometric Probability

11.5 Geometric Probability. By Sara Wojcak and Nicole Groves . Objectives. Solve problems involving geometric probability Solve problems involving sectors and segments of circles. Geometric Probability.

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11.5 Geometric Probability

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  1. 11.5 Geometric Probability By Sara Wojcak and Nicole Groves

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

  3. Geometric Probability • Geometric probability is probability that involves a geometric measure such as length or area. • You can find the probability that a point lies in a part of a two-dimensional figure by comparing the area of the part to the area of the whole figure. • P= area of part area of whole figure

  4. Key Concept A 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 When determining geometric probability with targets, we assume • That the object lands within the target area • It is equally likely that the object will land anywhere in the region B

  5. Example 1 • To find the probability of landing on the green portion of the shape, you will need to make gridlines in the shape. • Count the total amount of boxes. Then, count the number of green boxes. • 36 total boxes • 15 green boxes • So, the probability of landing on a green square is: 15 or 5 • 36 12

  6. Your Turn • Find the probability of landing on a blue space. • The area of the blue squares is 18 square units. • The entire area is 30 square units. • That means the probability of landing on a blue space is 18/30 or 9/15.

  7. Sectors and Segments of Circles • Sector- a region of a circle bounded by a central angle and its intercepted arc. • Find the area of a sector using the formula A=N r2 360 • Segment- the region of a circle bounded by an arc and a chord. Chord SectorArc Segment

  8. Example 2 • Find the area of the sector that is 46 degrees. a. Find the area of the sector A= N r2 360 Area of a sector = 46 (62) N=46, r=6 360 =4.6 Simplify b. Find the probability of a point landing within a certain region. ◦ P(46)= area of sector ◦ area of circle =4.6 .62 ◦ ~ 0.13 ~ The probability of it landing on 46 is .13 or 13% ◦ ◦ 57 46 80 70◦ 50 57

  9. Your Turn • Find the area of the indicated sector and the probability that a random point would be inside the sector. The diameter is 10 meters. • Find the area of the indicated sector using A= 90 52 360 A= 6.25 Find the probability that a random point lies in that sector using P=6.25 25 P= .25 or 25% 150◦ 40◦ 40◦ 90◦ 40◦

  10. Example 3 – Probability with segments • A regular hexagon is inscribed in a circle with a diameter of 14. • A. Find the area of the segment that is indicated. • Area of the sector: A= N r2 360 Area of a sector = 60 (72) N=46, r=6 360 =49 Simplify 6 ~ 25.66 ~

  11. Example 3 continued • Area of a triangle: Since the hexagon was inscribed in the circle, the triangle is 7 equilateral, with each side 7 units long. Use properties of a 30-60-90 3.5 triangle to find the apothem. The value of x is 3.5, the apothem is x√ 3 or 3.5√3 which is approximately 60◦ 7 7 6.06 30◦ Find the area of the triangle. A= 1 bh 2 = 1 (7)(6.06) 2 =6.06

  12. Example 3 continued (again) • Area of the segment: • Area of segment = area of sector- area of triangle Area of segment =25.66- 21.22 Substitution =4.44 Simplify • B. Find the probability that a point chosen at random lies in the indicated region. • Divide the area of the sector by the area of the circle to find the probability. First, find the area of the circle. The radius is 7, so the area is (72) or about 153.94 square units. P(indicated region)= area of segment area of circle = 4.44 153.94 = .03 The probability that a random point is on the indicated segment is about .03 or 3%

  13. Assignment • Page 625 #7- 25, 33

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