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.
+2862 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...
+118608 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 = \(\frac{2}{12.5}=16\%\)
