Luz Stokes

Answered

2022-07-16

A different approach to a common question

A unit stick is randomly broken into 3 pieces, it is given that these three pieces can make a triangle, what is the expected length of the medium-sized piece?

This a question we are all familiar with and anyone who has seen it anywhere/solved it knows that the expected length of the medium piece is $\frac{5}{18}$, we reach this conclusion by using $E(L+M+S)=1$, and we know we can calculate E(L) and E(S), so we just use $E(M)=1-E(L)-E(S)$ to get our answer, is there any way we can solely calculate the E(M) without calculating the other two values?

L: Length of the longest part

S: Length of the smallest part

M: Length of the medium part

A unit stick is randomly broken into 3 pieces, it is given that these three pieces can make a triangle, what is the expected length of the medium-sized piece?

This a question we are all familiar with and anyone who has seen it anywhere/solved it knows that the expected length of the medium piece is $\frac{5}{18}$, we reach this conclusion by using $E(L+M+S)=1$, and we know we can calculate E(L) and E(S), so we just use $E(M)=1-E(L)-E(S)$ to get our answer, is there any way we can solely calculate the E(M) without calculating the other two values?

L: Length of the longest part

S: Length of the smallest part

M: Length of the medium part

Answer & Explanation

minotaurafe

Expert

2022-07-17Added 22 answers

Step 1

An easy approach.

Let's arrange the segments from smallest to largest. Let the three segments be x, $x+y$ and $x+y+z$.

Now sum of all segments $=1$ $\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}3x+2y+z=1$;

with the conditions: $x\u2a7e0,y\u2a7e0,z\u2a7e0$ and $\text{and}x\u2a7d1/3,y\u2a7d1/2,z\u2a7d1$.

for normalisation, let n be a quantity $\u2a7d1/6$, then,

$\frac{x}{2}\u2a7d\frac{1}{6},\frac{y}{3}\u2a7d\frac{1}{6},\frac{z}{6}\u2a7d\frac{1}{6}$

x can be chosen randomly among a pool from [0,2n], y be chosen randomly from a pool [0,3n] and z from a pool [0,6n] where (x,y,z,n) $\in $ R.

Step 2

Therefore we can say:

Expected value of $x=n$, expected value of $y=1.5n$ and expected value of $z=3n$.

Expected length of middle segment = $x+y=2.5n$

Expected total length = $3x+2y+z=9n$

The the expected length of middle segment is

$(x+y)/(3x+2y+z)=2.5/9=5/18$

An easy approach.

Let's arrange the segments from smallest to largest. Let the three segments be x, $x+y$ and $x+y+z$.

Now sum of all segments $=1$ $\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}3x+2y+z=1$;

with the conditions: $x\u2a7e0,y\u2a7e0,z\u2a7e0$ and $\text{and}x\u2a7d1/3,y\u2a7d1/2,z\u2a7d1$.

for normalisation, let n be a quantity $\u2a7d1/6$, then,

$\frac{x}{2}\u2a7d\frac{1}{6},\frac{y}{3}\u2a7d\frac{1}{6},\frac{z}{6}\u2a7d\frac{1}{6}$

x can be chosen randomly among a pool from [0,2n], y be chosen randomly from a pool [0,3n] and z from a pool [0,6n] where (x,y,z,n) $\in $ R.

Step 2

Therefore we can say:

Expected value of $x=n$, expected value of $y=1.5n$ and expected value of $z=3n$.

Expected length of middle segment = $x+y=2.5n$

Expected total length = $3x+2y+z=9n$

The the expected length of middle segment is

$(x+y)/(3x+2y+z)=2.5/9=5/18$

Most Popular Questions