Random Angle II

Problem StatementPermalink

A right triangle is being formed with legs labeled A and B . The random lengths of legs
A and B are both IID Uniform(0,1) random variables.

Let θ be the angle (in radians) opposite to side A . Find the probability that:

$P(\theta > \pi/3)$

Original Problem Link: Click here

SolutionPermalink

Now we must find $P(\theta > \pi/3)$ .

Note that,

$\tan(\theta) = \frac{A}{B}$

where A and B both are IID Uniform(0,1) random variables.

For convenience, let’s fix B for now. Then, this problem becomes much simpler. It reduces to:

Given $\tan(\theta) = A/B$ , where A is Uniform(0,1) , B is a given constant,
find $P(\theta > \pi/3)$ .

Let’s try and solve this sub-problem:

$P(\theta > \pi/3) = P(\tan(\theta) > \sqrt{3}) = P(A/B > \sqrt{3}) = P(A > B\sqrt{3})$

Now, since A is a Uniform(0,1) random variable, clearly:

$P(A > x) = \max(0, 1-x)$

(How? 🤔)

If $x$ is between 0 and 1 , well and good.
If $x > 1$ , since A cannot be greater than 1, $P(A > x) = 0$ .
Still confused? Try intuitively finding $P(A>0)$ , $P(A>1/2)$ , and $P(A>1)$ . You will see the essence of the above expression.
Still confused? Read up on PDF (Probability Density Function) and CDF (Cumulative Density Function)
of the continuous Uniform distribution .

Therefore, for the smaller sub-problem above:

$P(A > B\sqrt{3}) = 1 - B\sqrt{3}$

for all $1 - B\sqrt{3} \geq 0 \Rightarrow B \leq 1/\sqrt{3}$ .

Essentially, this is equivalent to writing:

$P(\theta > \pi/3 \mid B) = 1 - B\sqrt{3}$

Why? 🤔
Because $P(\theta > \pi/3 \mid B)$ represents the probability of $\theta$ being greater than $\pi/3$ ,
given some B / conditioned on some B .

This is exactly what we calculated above! By fixing B , we found $P(\theta > \pi/3)$
in terms of some B , for a given B .

But of course, as the question states, B is also a Uniform(0,1) random variable.

We shall now use the fundamental law of total probability here.

Step 2: Using the Law of Total ProbabilityPermalink

Note that we can say:

$P(\theta > \pi/3) = \sum P(\theta > \pi/3 \mid B) P(B)$

for all B .

Now, we must compute $P(B)$ .

What exactly does this represent? 🤔

It basically asks:
“What is the probability of choosing a particular $B$ from $[0,1]$ ?”

Very small, right?

How do we quantify this? Using calculus!

We can say that the probability of choosing a particular $B$ from $[0,1]$ is essentially:

$\frac{dB}{1} = dB$

where dB is a very small term.

Notice that now our summation actually reduces to a simple integration , due to our usage of
“elemental B” or “dB” to model $P(B)$ .

Ideally, we should vary B from 0 to 1 , but note that B must be ≤ 1/√3 ,
otherwise $P(\theta > \pi/3 \mid B) = 0$ .

Therefore, we rewrite the above expression as a definite integral:

$P(\theta > \pi/3) = \int_{0}^{1/\sqrt{3}} (1 - B\sqrt{3}) dB$

Step 3: Solving the IntegralPermalink

Expanding the integral:

$\int_{0}^{1/\sqrt{3}} (1 - B\sqrt{3}) dB$

We solve separately:

$\int 1 \, dB = B$

$\int B\sqrt{3} \, dB = \frac{\sqrt{3} B^2}{2}$

Evaluating from 0 to 1/√3 :

$P(\theta > \pi/3) = \left[ B - \frac{\sqrt{3} B^2}{2} \right]_{0}^{1/\sqrt{3}}$

Substituting B = 1/√3 :

$P(\theta > \pi/3) = \frac{1}{\sqrt{3}} - \frac{\sqrt{3} (1/3)}{2}$

$= \frac{1}{\sqrt{3}} - \frac{\sqrt{3}}{6}$

$= \frac{2}{2\sqrt{3}} - \frac{1}{2\sqrt{3}}$

$= \frac{1}{2\sqrt{3}}$

Thus,

$\boxed{P(\theta > \pi/3) = \frac{1}{2\sqrt{3}}}$

ConclusionPermalink

Using a combination of probability concepts, the law of total probability, and calculus ,
we found that the probability of θ exceeding π/3 is 1/2√3 .

This problem highlights how fundamental probability techniques can be used to analyze geometric randomness .

💡 Did you enjoy this problem? Check out more puzzles in the Problems section!