11.5 Geometric Probability

Austin Varghese and Lane Driskill

Objectives

• Solve problems involving geometric probability

• Solve problems involving sectors and segments of circles

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

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, isP(B)= area of region B/ area of region A

A

B

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

Example #1

• Find the probability that a point chosen at random will lie in ∆MNC

2x

x

x

Example #1 (cont.)

(2x) · (2x)= 4x² Area of Square½x² Area of ∆MNC½x² ÷ 4x² P(point in the ∆)=½x² · ¼x² Division=⅛ Multiplication

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, thenA= N/360 πr²

N°

r

Example #2

• Find the area of the green area. Assume the diameter of the circle is 12 inches

40°35°

75° 210°

Example #2 (cont.)

A=N/360(π)(r²) Area of a sector =35/360(π)(6²) N=35, r=6=3.5π Simplify

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

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

Example #3 (cont.)

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

Example #3 (cont.)

b. Find the area of the triangleA=½bh Area of a triangleA=½(5)(5) b=5, h=5=12.5 Simplifyc. Find the area of the segment≈19.63 – 12.5 Substitution≈7.13 Simplify

Example #3 (cont.)

• Part 2: Find the probabilityP(red)= area of segment/area of circle≈7.13/78.54 Substitution≈0.09 SimplifyAnswer: 0.09 or 9% chance that a random point would land in the red segment

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

Example #4 (cont.)

A=(N/360) · 100 Finding probability without diameter

A=(226/360) · 100 SimplifyA≈62.77 Solve

Answer: There is a 63% that a dog lover will be chosen at random

Assignment

• Pre-AP Geometry- pg 625-626 #7-23 all