Suppose we have a stick of length 1.a) We randomly uniformly choose a point and break the stick into two pieces. Find the expected length of the smaller piece. b) We randomly uniformly choose two points (independently) and break thestick into three pieces. Find the probability that the three resulting piecescan be arranged to form a triangle (i. e. all triangle inequalities are satisfied;i. e no piece is longer than the sum of the other two).

Step-by-step explanation:

1) The smaller sticks will range in length from almost 0 unit up to a maximum of 0.5 unit, with each length equally possible.

Therefore, the average length will be about (0 + 0.5)/2 = 0.25 unit

2)If you assume that each break in the stick is uniformly distributed along the length of the stick and is independent of the location of the other break, then the odds are 25% that you will be able to form a triangle with the 3 pieces.

We'll call the length of the stick 1, so each break can occur at a position in the interval [0,1]. Let x and y represent the two breaks. Then we can look at the area of the region in the square bounded x=0, x=1, y=0, y=1, which represents combinations of x and y, for which we can form a triangle. Since the area of the whole square is 1, the area of the region inside is our probability.

If y>x, then the lengths of the pieces are x, y-x, and 1-y.

The triangle inequality must hold for each combination of edges.

for y>x ...

x+y−x≥1−y

x+1−y≥y−x

y−x+1−y≥x

these simplify to...

for y>x ...

y≥1/2

x+1/2≥y

x≤1/2

If we cut our 1x1 square into two triangles along the line x=y,

then the region in the upper triangle which satisfies the inequalities above forms a smaller triangle which connects the midpoints of the upper triangle.

The lower triangle (x>y), is just a reflection about x=y of the upper triangle, so together, the entire region looks like a bow-tie at a 45 degree angle.

This region takes up 25% of the square, so the probability that you can form a triangle is 25%