# Probability that Three Pieces Form a Triangle

## Problem 754

We have a stick of a unit length. Two points on the stick will be selected randomly (uniformly along the length of the stick) and independently. Then we break the stick at these two points so that we get three pieces of the stick. What is the probability that these three pieces form a triangle?

## Solution.

Let us call the left end of the stick the origin. Let $X$ be the length from the origin to the first selected point. Let $Y$ be the length from the origin to the second selected point.

Note that when $X=Y$, we have only two pieces and they cannot form a triangle, so we assume $X \neq Y$.
We first assume that $X < Y$. Then after breaking the stick, we have three pieces of length $X$, $Y-X$, and $1-Y$, respectively.

Now, let us consider a condition so that three segments of length $a$, $b$, and $c$ form a triangle in general. It is necessary and sufficient that the sum of the lengths of any two segments is bigger than the length of the rest to form a triangle.
Namely, we must satisfy all of the following inequalities.
\begin{align*}
a + b &> c\\
b + c &> a\\
c + a &> b.
\end{align*}

Applying this condition to the current situation, we must have
\begin{align*}
X + (Y-X) &> 1-Y\\
(Y-X) + (1-Y) & > X\\
(1- Y) + X &> Y – X.
\end{align*}
Simplifying these, we obtain
\begin{align*}
Y &> \frac{1}{2}\\
X &< \frac{1}{2}\\ Y &< X + \frac{1}{2}. \end{align*} The region in the unit square satisfying these inequality is depicted in the figure below (orange region).

Note that the joint density function of $X$ and $Y$ are uniformly distributed on the unit square. Thus, the probability that three pieces form a triangle under the assumption $X < Y$ is given by the area of the orange triangular region, which is $1/8$. Note that by symmetry, the case $X > Y$ gives the same probability. Hence, the desired probability is
\begin{align*}
&P(\text{form a triangle}) \\
&= P(\text{form a triangle} \mid X < Y) + P(\text{form a triangle} \mid X > Y)\\
&\frac{1}{8}+\frac{1}{8} = \frac{1}{4}.
\end{align*}

Alternatively, we can find the probability for the case $X > Y$ as follows. By symmetry argument, we simply need to exchange $X$ and $Y$ and obtain the conditions
\begin{align*}
X &> \frac{1}{2}\\
Y &< \frac{1}{2}\\ X &< Y + \frac{1}{2}. \end{align*} The blue region below shows the area surrounded by these inequalities. The area of this blue region is $1/8$. Thus, the total area is $1/8 + 1/8 = 1/4$ as before.

### More from my site

• Linearity of Expectations E(X+Y) = E(X) + E(Y) Let $X, Y$ be discrete random variables. Prove the linearity of expectations described as $E(X+Y) = E(X) + E(Y).$ Solution. The joint probability mass function of the discrete random variables $X$ and $Y$ is defined by $p(x, y) = P(X=x, Y=y).$ Note that the […]
• Probability that Alice Tossed a Coin Three Times If Alice and Bob Tossed Totally 7 Times Alice tossed a fair coin until a head occurred. Then Bob tossed the coin until a head occurred. Suppose that the total number of tosses for Alice and Bob was $7$. Assuming that each toss is independent of each other, what is the probability that Alice tossed the coin exactly three […]
• Independent Events of Playing Cards A card is chosen randomly from a deck of the standard 52 playing cards. Let $E$ be the event that the selected card is a king and let $F$ be the event that it is a heart. Prove or disprove that the events $E$ and $F$ are independent. Definition of Independence Events […]
• Conditional Probability When the Sum of Two Geometric Random Variables Are Known Let $X$ and $Y$ be geometric random variables with parameter $p$, with $0 \leq p \leq 1$. Assume that $X$ and $Y$ are independent. Let $n$ be an integer greater than $1$. Let $k$ be a natural number with $k\leq n$. Then prove the formula \[P(X=k \mid X + Y = n) = […]
• If a Smartphone is Defective, Which Factory Made It? A certain model of smartphone is manufactured by three factories A, B, and C. Factories A, B, and C produce $60\%$, $25\%$, and $15\%$ of the smartphones, respectively. Suppose that their defective rates are $5\%$, $2\%$, and $7\%$, respectively. If a smartphone of this model is […]
• Probability of Having Lung Cancer For Smokers Let $C$ be the event that a randomly chosen person has lung cancer. Let $S$ be the event of a person being a smoker. Suppose that 10% of the population has lung cancer and 20% of the population are smokers. Also, suppose that we know that 70% of all people who have lung cancer […]
• Jewelry Company Quality Test Failure Probability A jewelry company requires for its products to pass three tests before they are sold at stores. For gold rings, 90 % passes the first test, 85 % passes the second test, and 80 % passes the third test. If a product fails any test, the product is thrown away and it will not take the […]
• Pick Two Balls from a Box, What is the Probability Both are Red? There are three blue balls and two red balls in a box. When we randomly pick two balls out of the box without replacement, what is the probability that both of the balls are red? Solution. Let $R_1$ be the event that the first ball is red and $R_2$ be the event that the […]

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

##### Upper Bound of the Variance When a Random Variable is Bounded

Let $c$ be a fixed positive number. Let $X$ be a random variable that takes values only between $0$ and...

Close