Section 2.2 Exercise 12

Take a stick of unit length and break it into three pieces, choosing the break points at random. (The break points are assumed to be chosen simultaneously.) What is the probability that the three pieces can be used to form a triangle? Hint: The sum of the lengths of any two pieces must exceed the length of the third, so each piece must have length < 1/2.

Solution

If any piece is longer than 1/2, then the triangle cannot be formed. Let \(B\) and \(C\) be two break points - \(0<B<1\) and \(0<C<1\). Let us assume that the first piece is longer than 1/2. Probability of that is \(P(0.5<B<1) \times P(0.5<C<1) = 0.5 * 0.5 = 0.25\). Similarly, we can assume that the second piece is longer than 1/2 and then that the third piece is longer than 1/2. Probability of each is the same, so probability of one of three pieces being longer than 1/2 is \(3 * 0.25 = 0.75\). Probability of none of the pieces being longer than 1/2 is \(1-0.75 = 0.25\).

I have confirmed this with an R code.

# Initialize variables
n <- 100000
B <- runif(n)
C <- runif(n)
p <- rep(0,n)

# Loop through all samples
for (i in 1:n) {
  if (B[i]>C[i]) {
    t <- C[i]
    C[i] <- B[i]
    B[i] <- t
  }
  if ((B[i]<.5) & ((C[i]-B[i])<.5) & ((1-C[i])<.5)) {
    p[i] <- 1
  }
}

cat("Probability:",sum(p)/n)
## Probability: 0.2484

Section 2.2 Exercise 13

Take a stick of unit length and break it into two pieces, choosing the break point at random. Now break the longer of the two pieces at a random point. What is the probability that the three pieces can be used to form a triangle?

Solution

R simulation is below.

# Initialize variables
n <- 100000
B <- runif(n)
p <- rep(0,n)

# Loop through all samples
for (i in 1:n) {
  if (B[i]<0.5) {
    C <- runif(1,B[i],1)
    b <- B[i]
  } else {
    C <- B[i]
    b <- runif(1,0,B[i])
  }
  if ((b<.5) & ((C-b)<.5) & ((1-C)<.5)) {
    p[i] <- 1
  }
}

cat("Probability:",sum(p)/n)
## Probability: 0.38787