Transcript Geometry - BakerMath.org

Geometry
Geometric Probability
Goals
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July 17, 2015
Know what probability is.
Use areas of geometric figures to
determine probabilities.
Probability
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July 17, 2015
A number from 0 to 1 that
represents the chance that an event
will occur.
P(E) means “the probability of event
E occuring”.
P(E) = 0 means it’s impossible.
P(E) = 1 means it’s certain.
P(E) may be given as a fraction,
decimal, or percent.
Probability
Number of Successful Outcomes
P(E)=
Total number of Outcomes
Example
A ball is drawn at random from the box.
What is the probability it is red?
?
2
P(red) =
?
9
July 17, 2015
Probability
Number of Successful Outcomes
P(E)=
Total number of Outcomes
A ball is drawn at random from the box.
What is the probability it is green or black?
?
3
P(green or black) =
?
9
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Probability
Number of Successful Outcomes
P(E)=
Total number of Outcomes
A ball is drawn at random from the box.
What is the probability it is green or black?
1
P(green or black) =
3
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Geometric Probability
Based on lengths of segments and
areas of figures.
Random:
Without plan or order. There is no
bias.
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Probability and Length
Let AB be a segment that contains the
segment CD. If a point K on AB is
chosen at random, then the probability
that it is on CD is
Length of CD
P(K is on CD) 
Length of AB
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Example 1
Find the probability that a point
chosen at random on RS is on JK.
R
J
K
S
1 2 3 4 5 6 7 8 9 10 11 12
JK = 3
RS = 9
Probability = 1/3
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Your Turn
Find the probability that a point chosen at
random on AZ is on the indicated segment.
A
B
C
D
E Z
1 2 3 4 5 6 7 8 9 10 11 12
AB  1
5
AC  2
5
BD  1
2
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AE  9
10
EZ  1
10
BZ  4
5
Probability and Area
Let J be a region that contains
region M. If a point K in J is chosen
at random, then the probability that
it is in region M is
Area of M
P(K is in M)=
Area of J
M
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J
K
Example 2
Find the probability that a randomly
chosen point in the figure lies in the
shaded region.
8
8
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Example 2 Solution
Area of Square = 82 = 64
8
8
Area of Triangle
A=(8)(8)/2 = 32
Area of shaded region
8
64 – 32 = 32
Probability:
32/64 = 1/2
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Example 3
Find the probability that a randomly
chosen point in the figure lies in the
shaded region.
5
July 17, 2015
Example 3 Solution
Area of larger circle
10
5
5
A = (102) = 100
Area of one smaller circle
A = (52) = 25
Area of two smaller circles
Probability
50
1

100 2
July 17, 2015
A = 50
Shaded Area
A = 100 - 50 = 50
Your Turn
A regular hexagon is inscribed in a circle.
Find the probability that a randomly chosen
point in the circle lies in the shaded region.
6
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Solution
Find the area of the hexagon:
6 6
?
?
3
?
33 33
1
A  ap
2
1

3 3 36 
2


 54 3  93.53
July 17, 2015
Solution
19.57
93.53
6 6
113.1
3
3
33 3
Find the area of the circle:
A = r2
A=36  113.1
Shaded Area
Circle Area – Hexagon Area
113.1 – 93.63 =19.57
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Solution
19.57
Probability:
Shaded Area ÷ Total Area
19.57/113.1 = 0.173
6 6
113.1
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3
3
33 3
17.3%
Example 4
If 20 darts are randomly thrown at
the target, how many would be
expected to hit the red zone?
10
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Example 4 Solution
Radius of small circles:
5
Area of one small circle:
25
Area of 5 small circles:
125
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10
Example 4 Solution continued
Radius of large circle:
15
Area of large circle:
(152) = 225
Red Area:
(Large circle – 5 circles)
225  125 = 100
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10
10
5
Example 4 Solution continued
Red Area:100
Total Area: 225
10
Probability:
100
4
  0.444...
225
9
This is the probability for each dart.
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Example 4 Solution continued
Probability:
100
4
  0.444...
225
9
For 20 darts, 44.44% would
likely hit the red area.
20  44.44%  8.89, or about
9 darts.
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10
Your Turn
500 points are randomly selected in
the figure. How many would likely be
in the green area?
5 3
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Solution
500 points are randomly selected in the figure.
How many would likely be in the green area?
Area of Hexagon:
A = ½ ap
A = ½ (53)(60)
10
5 3
60
5
Area of Circle:
A = r2
10
July 17, 2015
A = 259.81
A = (53)2
A= 235.62
Solution
500 points are randomly selected in the figure.
How many would likely be in the green area?
Area of Hexagon:
A = 259.81
Area of Circle:
5 3
A= 235.62
Green Area:
259.81 – 235.62
24.19
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Solution
500 points are randomly selected in the figure.
How many would likely be in the green area?
Area of Hexagon:
A = 259.81
Green Area:
5 3
24.19
Probability:
24.19/259.81 =
0.093 or 9.3%
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Solution
500 points are randomly selected in the figure.
How many would likely be in the green area?
Probability:
0.093 or 9.3%
For 500 points:
5 3
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500  .093 = 46.5
47 points should
be in the green
area.
Summary
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July 17, 2015
Geometric probabilities are a ratio
of the length of two segments or a
ratio of two areas.
Probabilities must be between 0 and
1 and can be given as a fraction,
percent, or decimal.
Remember the ratio compares the
successful area with the total area.
Practice Problems
July 17, 2015