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(1) 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?

(2) 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 you can help, that's great! If I solve it before, I'll explain it on here so others can learn too!

THANKS SO MUCH!

Jeff123 Sep 26, 2017

#2**+1 **

(1) 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?

I stuffed up on question 1 but it is very similar to question 2

**Since you have never bothered to respond to my answers there is not point in me doing it properly now.**

Melody Sep 26, 2017

#4**+1 **

I found an old question that was a bit like this, I remember Bertie helped me with it.

https://web2.0calc.com/questions/a-line-segment-of-length-5-is-broken-at-two-random

Melody Sep 26, 2017

#6**+2 **

QUESTION 2

(2) 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?

I drew this picture in Geogebra. The real thing is interactive and by playing with it I concluded that if all the angles of the triangle are acute then the centre will be included.

Now it is very similar to the first problem only maybe easier.

Let the 3 angles be x, 180-y and y-x

Where y>x and each of thes angles is acute.

\(0 x \qquad (3)\\ x<90\qquad (4)\\ 180-y<90\\ -y<-90\\y>90 \qquad (5)\\ y-x<90\\ y

Here is the Boolean contour map so if y>x then the probability is 1/8

BUT if x>y then you have the inverse of this region. I mean this region reflected over the line y=x

SO

**It seems to me that the probability that the centre will be in the triangle is 1/4**

Melody Sep 27, 2017

#7**0 **

Jeff123,

It would be really nice if you would comment on my answers. I think my answers are correct but this technique is quite new to me. If you have any reason to think the answers are incorrect (like maybe the probability answer is printed in your text book) it would be really nice if you share what you know.

Melody Sep 27, 2017