A stick has a length of units. The stick is then broken at two points, chosen at random. What is the probability that all three resulting pieces are shorter than units?
I assumed " shorter than 3 units":
I will assume two cut points x and y, each chosen using a uniform distribution of reals from [0,5]. Let us consider the square in 2 -space with corners at (0,0) and (5,5) and find the areas within that square for which the longest piece has length ≥3.
That occurs when both x and y are ≤2. That corresponds to a 2×2 square in the lower left. Similarly, when both are ≥3, we have an area 4 square in the upper right.
It also occurs when x−y≥3. That corresponds to a triangle in the lower right bounded by the line from (3,0) to (5,2). It has area 2. For y−x≥3, we have an area 2 triangle at the upper left.
Those four areas are disjoint and add up to total area of 12. In the area 13 space that remains, no piece has length ≥3.
Answer: 13/25
+118690 A stick has a length of units. The stick is then broken at two points, chosen at random. What is the probability that all three resulting pieces are shorter than units?
First use contour mapping to graph the sample space.
Let the lengths be x, y-x and 5-y
they add up to 5.
I know that each of thise is between 0 and 5 units
0
0
0
x
y>x (2)
0<5-y<5
-5<-y<0
0
Here is your sample space. Every possible outcome is in there.
Now you have to graph the favourable sample space. LOL you did not proof read your question!
You want people to answer you. You even tell an answerer that their answer is wrong.
Of course it is wrong, there is no question. What a waste of our time.
