Non-geometric way to calculate expected value of breaks?"A bar is broken at random in two...

Elisabeth Esparza

Elisabeth Esparza

Answered

2022-07-19

Non-geometric way to calculate expected value of breaks?
"A bar is broken at random in two places. Find the average size of the smallest, of the middle-sized, and of the largest pieces."
The author gives what seems like a complicated geometric way of calculating the probabilities. He arrives at the solutions 1/9, 5/18, and 11/18. Is there a simpler, non-geometric way of calculating these probabilities?

Answer & Explanation

Damarion Pierce

Damarion Pierce

Expert

2022-07-20Added 11 answers

Step 1
Here's a sketch. Pick points 0 x y 1. Now let's find the expected value of x given that the interval [0,x] is shortest. That is, we look at the region satisfying x y x y x x 1 y
You can draw your own pictures. We then get a triangle with 0 x 1 / 3, 2 x y 1 x. This triangle has area 1/6 (by inspection) or by double-integration, and 0 1 / 3 2 x 1 x x d y d x = 1 54 , so the expected value of x is 1 54 1 6 = 1 9 .
Step 2
Next case. Suppose the interval [0,x] has middle length. This corresponds to two regions:
x y y x x x 1 y
or x y x y x 1 y x
Interestingly, these both have area 1/12 and for each of these we get x d A = 5 216 . So the expected value of x in the middle case is 5 216 1 12 = 5 18
Lillie Pittman

Lillie Pittman

Expert

2022-07-21Added 4 answers

Step 1
For the k t h smallest piece indicated by S ( k ) , the expected length is given by
E ( n S ( k ) ) = E ( X ( k ) ) = i = 0 k 1 1 n i
which simplifies to the formula given in the book of 50 challenging problems.
Step 2
For the question asked, n = 3 and calculate the sums for k = 1 , 2 , 3, which we find to be
E ( S ( 1 ) ) = 1 9 E ( S ( 2 ) ) = 5 18 E ( S ( 3 ) ) = 11 18

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