Solve problems involving geometric
probability
 Solve problems involving sectors and
segments of circles

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

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
A
B
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
 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




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.

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
Sector
Arc
Segment

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
◦
57 ◦
46
80
◦
70◦
~
The probability of it landing on 46 is .13 or 13%
◦
57
50 ◦

Find the area of the indicated
sector and the probability that
a random point would be
inside the sector. The diameter
is 10 meters.
150◦
40◦
40◦
40◦
90◦

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%

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
= 60 (72)
360
=49
6
~ 25.66
~
Area of a sector
N=46, r=6
Simplify

Area of a triangle:
Since the hexagon was inscribed in the circle, the triangle is
equilateral, with each side 7 units long. Use properties of a 30-60-90
triangle to find the apothem.
The value of x is 3.5, the apothem is x√ 3 or 3.5√3 which is approximately
6.06
7
3.5
60◦
7
30◦
Find the area of the triangle.
A= 1 bh
2
= 1 (7)(6.06)
2
=6.06
7


Area of the segment:
› Area of segment = area of sector- area of triangle
= 25.66- 21.22 Substitution
Area of segment
=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%

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