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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!

 Sep 26, 2017
 #1
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HELP PLEASE

 Sep 26, 2017
 #2
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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?

 

 

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.

 Sep 26, 2017
edited by Melody  Sep 26, 2017
edited by Melody  Sep 26, 2017
edited by Melody  Sep 26, 2017
edited by Melody  Oct 7, 2017
edited by Melody  Oct 7, 2017
 #4
avatar+118687 
+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

 Sep 26, 2017
 #6
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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

 Sep 27, 2017
edited by Melody  Sep 27, 2017
edited by Melody  Oct 7, 2017
 #8
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Once again it is not displaying properly only this time it is the LaTex that is not displaying correctly  sad

Melody  Sep 27, 2017
 #7
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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. 

 Sep 27, 2017

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