Problem Statement

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

---

Solution

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 Probability

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 Integral

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,

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

---

Conclusion

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!