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A line segment is broken at two random points along its length. What is the probability that the three new segments can be arranged to form a triangle?

Three points are selected randomly on the circumference of a circle. What is the probability that the triangle formed by these three points contains the center of the circle?

Apr 17, 2019

#1
+2

The first question has been answered here many time before. If you search it in google you will find it.

Both questions are most easlily solved with the use of probability contour maps.

Three points are selected randomly on the circumference of a circle. What is the probability that the triangle formed by these three points contains the center of the circle?

If it is a right angled triangle then the diameter will be one sides and the centre will be on this side.

If the triangle is a acute angled then the centre will be included.

If the triangle is obtuse then the centre will not be included.

I am actually going to look at the set of obtuse angled trianges which is mutually exclusive to the set of ones that include the centre.

Let the angles be x, y, and 180-(x+y)

The set of all possible triangles is contained in the region

x>0    y>0    x+y<180

Sample space area = 0.5* 180*180 units squared. Now it will be obtuse if  x>90. or   y>90   or    x+y<90 So the probability of being acute = probability of the centre being included in the triangle.

Are of included region =  0.5*90*90

$$\text{P(centre included) = }\frac{0.5*90*90}{0.5*180*180}=\frac{1}{4}$$

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Apr 17, 2019

#1
+2

The first question has been answered here many time before. If you search it in google you will find it.

Both questions are most easlily solved with the use of probability contour maps.

Three points are selected randomly on the circumference of a circle. What is the probability that the triangle formed by these three points contains the center of the circle?

If it is a right angled triangle then the diameter will be one sides and the centre will be on this side.

If the triangle is a acute angled then the centre will be included.

If the triangle is obtuse then the centre will not be included.

I am actually going to look at the set of obtuse angled trianges which is mutually exclusive to the set of ones that include the centre.

Let the angles be x, y, and 180-(x+y)

The set of all possible triangles is contained in the region

x>0    y>0    x+y<180

Sample space area = 0.5* 180*180 units squared. Now it will be obtuse if  x>90. or   y>90   or    x+y<90 So the probability of being acute = probability of the centre being included in the triangle.

Are of included region =  0.5*90*90

$$\text{P(centre included) = }\frac{0.5*90*90}{0.5*180*180}=\frac{1}{4}$$

Melody Apr 17, 2019
#2
+1

That's a nice approach, Melody   !!!   CPhill  Apr 17, 2019
#4
+1

Thanks Chris and guest. :)

Melody  Apr 18, 2019