More help would be greatly appreciated. On average, how long are the pieces that contain the 6-inch mark?

Question

A Manufacturing Company makes all sorts of mathematical tools: compasses, protractors, slide rules — you name it!

Recently, there was an issue with the production of foot-long rulers. It seems that each ruler was accidentally sliced at three random points along the ruler, resulting in four pieces.

Looking on the bright side, that means there are now four times as many rulers — they just happen to have different lengths. On average, how long are the pieces that contain the 6-inch mark?

Answer 1

We will use I[a,b] for the integral from a to b and E[a,b] for the expectation from a to b

The probability that there are 0, 1, 2, and 3 cuts to the left of the 6-inch mark are 1/8, 3/8, 3/8, and 1/8

As P(all 3 >= x) = (12-x)^3/1728, f(x) = 3(12-x)^2/1728. Conditioning this on all three being >= 6, f(x) = 3(12-x)^2*8/1728 = (12-x)^2/72

Then, the average length is I[6,12] x(12-x)^2/72 dx = I[6,12] 2x - x^2/3 + x^3/72 dx = x^2 - x^3/9 + x^4/288 E[6,12] = 12^2 - 12^3/9 + 12^4/288 - (6^2 -6^3/9+6^4/288) = (144 - 192 + 72) - (36 - 24 + 9/2) = 7 1/2

If all 3 <= x, we will have the same average value of 7 1/2

Next, consider 2 to the left of 6 and 1 to the right of 6. As the 1 will be uniformly distributed from 6 to 12, it will average 9 or 3 to the right of 6.

For the 2 to the left of 6, P(both are less than or equal to x) is x^2/36, so the density is 2x/36 = x/18, and the average location of the second is I0, 6] x^2/18 dx= x^3/54 E[0, 6] = 6^3/54 - 0 = 4. Then, the average amount to the left of 6 is 6 - 4 = 2.

Thus, the length is 2 + 3 = 5

By symmetry, the average length for 1 to the left of 6 and 2 to the right of 6 will also be 5.

Thus, the expected length is 7 1/2 * 1/8 + 5 * 3/8 + 5 * 3/8 + 7 1/2 * 1/8 = 15/16 + 15/8 + 15/8 + 15/16 = 45/8