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A stick is 5 units long. It is broken at two points, chosen at random. What is the probability that all three pieces are longer than 1 unit?

 

I tried to set variables at different points. And I know that these variables range over an interval. Variables are x, y, and z.

 Feb 14, 2020
 #1
avatar+2646 
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Two conditions:

 

probability of first break that will create two sticks greater than 1

 

Break must be greater than 1 OR less than 4

 

Probability that second break will create three sticks greater than 1

 

second break must be greater than 2 OR less than 3

 

not sure how am I supposed to calculate for probabilty... its always been my weakness...

 
 Feb 14, 2020
 #2
avatar+107414 
+1

A stick is 5 units long. It is broken at two points, chosen at random. What is the probability that all three pieces are longer than 1 unit?

 

I will do this with a probability contour map.

Let the 3 pieces be x, y and 5 - (x+y) units long.

so

\(1) \;0 x+y\;\;  and\;\;  x+y>0\\ 5-x>y  \;\;  and\;\;  y>-x\\\)

3)  \( y<-x+5\;\;  \cap \;\; y>-x\)

 

 I will graph these and it will represent the sample space.

Area of sample space = 0.5*5*5 = 12.5units squared.

 

Now for the desired area.

What is the probability that all three pieces are longer than 1 unit?
\(x>1, \;\; y>1\\ and\\ 5-(x+y)>1\\ 4>y+x\\ 4-x>y\\ y<-x+4\\ \)

 

 

Area of middle triangle (desired outcome) = 2units squared

 

 

What is the probability that all three pieces are longer than 1 unit  =  \(\frac{2}{12.5}=16\%\)

.
 
 Feb 14, 2020

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