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11.5 Geometric Probability Austin Varghese and Lane Driskill.

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Presentation on theme: "11.5 Geometric Probability Austin Varghese and Lane Driskill."— Presentation transcript:

1 11.5 Geometric Probability Austin Varghese and Lane Driskill

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

3 Key Vocabulary Words Geometric probability- the probability that involves a geometric measure such as length or area Sector- a region of a circle bounded by a central angle and its intercepted arc Segment- the region of a circle bounded by an arc and a chord

4 Probability and 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

5 Things that you have to know When determining geometric probability with targets, we assume: that the object lands within the target area, and it is equally likely that the object will land anywhere in the region

6 Example #1 Find the probability that a point chosen at random will lie in ∆MNC 2x x x

7 Example #1 (cont.) (2x) · (2x)= 4x²Area of Square ½x²Area of ∆MNC ½x² ÷ 4x²P(point in the ∆) =½x² · ¼x²Division =⅛Multiplication

8 Area of a Sector If a sector of a circle has an area of A square units, a central angle measuring N˚, and a radius of r units, then A= N/360 πr² N° r

9 Example #2 Find the area of the green area. Assume the diameter of the circle is 12 inches 210°

10 Example #2 (cont.) A=N/360(π)(r²)Area of a sector =35/360(π)(6²)N=35, r=6 =3.5πSimplify

11 Probability with Segments To find the area of a segment, subtract the area of the triangle formed by the radii and the chord from the area of the sector containing the segment

12 Example #3 Find the probability that a random point will lie in the red segment R A regular square is inscribed in circle R with a diameter of 10 10

13 Example #3 (cont.) a. Find the area of the red segment A=N/360(π)(r²)Area of sector A=90/360 (π)(5²)N=90, r=5 =25/4(π)Simplify ≈19.63Use a calculator

14 Example #3 (cont.) b. Find the area of the triangle A=½bhArea of a triangle A=½(5)(5)b=5, h=5 =12.5Simplify c. Find the area of the segment ≈19.63 – 12.5Substitution ≈7.13Simplify

15 Example #3 (cont.) Part 2: Find the probability P(red)= area of segment/area of circle ≈7.13/78.54Substitution ≈0.09Simplify Answer: 0.09 or 9% chance that a random point would land in the red segment

16 Example #4 If a person is chosen at random from the groups, find the probability of a dog lover being chosen. 226° 94° 40° What’s your favorite pet? dogs cats rabbits

17 Example #4 (cont.) A=(N/360) · 100Finding probability without diameter A=(226/360) · 100Simplify A≈62.77Solve Answer: There is a 63% that a dog lover will be chosen at random

18 Assignment Pre-AP Geometry- pg 625-626 #7-23 all


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