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 .
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