McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
3. Geometric Probability
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Exercise 25 Page 903

In geometric probability, points on a segment or in a region of a plane represent outcomes. The geometric probability of an event is a ratio that involves geometric measures such as length or area.

0.535 or 53.5 %

Practice makes perfect

We can use geometric models to solve certain types of probability problems. In geometric probability, points on a segment or in a region of a plane represent outcomes. The geometric probability of an event is a ratio that involves geometric measures such as length or area. Consider the given diagram.

We are told that a point in the figure is chosen at random, and want to find the probability that the point lies in the shaded region. The probability that the point is in the shaded region is the ratio of the area of the shaded region to the area of the figure. P(The point is in the shaded region)= [0.8em] Area of the shaded region/Area of the figure We will find the area of the shaded region and the area of the entire figure one at a time. Then, we will find their ratio.

Area of the Figure

The figure is a circle with a diameter of 5 units. Since the diameter equals 5 units, the radius of the circle is 52= 2.5. Let's substitute this value for r in the formula for the area of a circle.

A_C=Ï€ r^2
A_C=Ï€ ( 2.5^2)
â–¼
Evaluate right-hand side
A_C=Ï€ (6.25)
A_C=19.634954...
A_C≈19.63

The area of the circle is equal to about 19.63. Area of the Figure: 19.63

Area of the Shaded Region

Notice that the area of the shaded region is the difference between the area of a circle and the area of two right triangles. Let's first focus on the smaller triangle!

Since the radius of the circle is equal to 2.5, we know that both of the legs of the smaller triangle also measure 2.5. Now, we can substitute b= 2.5 and h= 2.5 into the formula for the area of a triangle.

A_S=1/2bh
A_S=1/2( 2.5)( 2.5)
â–¼
Simplify right-hand side
A_S=1/2(6.25)
A_S=6.25/2
A_S=3.125

The area of the smaller triangle is equal to 3.125. Now, we can focus on the larger triangle!

Since the triangle is a right triangle, we can use the Pythagorean Theorem to find the length of the missing side of the triangle. The hypotenuse is the side opposite the right angle, for this triangle it measures 5 units. The two other sides are the legs, 4 and a. We can now form an equation with the Pythagorean Theorem, that can be solved for a.

a^2+h_a^2=c^2
a^2+ 4^2= 5^2
â–¼
Solve for a
a^2+16=25
a^2=25-16
a^2=9
a=±sqrt(9)
a=±3

a > 0

a=3

Now, we can substitute a= 3 and h_a= 4 into the formula for the area of a triangle.

A_L=1/2ah_a
A_L=1/2( 3)( 4)
â–¼
Simplify right-hand side
A_L=1/2(12)
A_L=12/2
A_L=6

The area of the larger triangle is 6. To find the area of the shaded region, which is formed by the difference between the circle and two triangles, we need to subtract A_S and A_L from A_C.

A=A_C-A_S-A_L
A=19.63-3.125-6
â–¼
Simplify right-hand side
A=10.505
A≈10.51

The shaded area is equal to about 10.51. Shaded Area: 10.51

Probability

As previously mentioned, the probability that the point is in the shaded region is the ratio of the area of the shaded region to the area of the figure. Since we already know both areas, we can find their ratio.

P=Area of the shaded region/Area of the figure
P=10.51/19.63
â–¼
Simplify right-hand side
P=0.535405...
P≈ 0.535
P≈ 53.5 %

The probability that a point from the figure chosen at random is in the shaded region is about 0.535, which can be also written as 53.5 %.