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

\[P(\theta > \pi/3)\]

Original Problem Link: Click here

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

Let’s try and solve this sub-problem:

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

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.

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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 ≤ \(\frac{1}{\sqrt 3}\) ,
otherwise \(P(\theta > \pi/3 \mid B) = 0\) .

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

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

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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,

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

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Conclusion

Using a combination of probability concepts, the law of total probability, and calculus ,
we found that the probability of \(θ\) exceeding \(\frac{\pi}{3}\) is \(\frac{1}{2√3}\) .

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

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