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.

Guest Feb 14, 2020

#1**+2 **

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...

CalculatorUser Feb 14, 2020

#2**+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 = \(\frac{2}{12.5}=16\%\)

.Melody Feb 14, 2020