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(θ>π/3)Original Problem Link: Click here
SolutionPermalink
Now we must find P(θ>π/3).
Note that,
tan(θ)=ABwhere 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(θ)=A/B, where A is Uniform(0,1) , B is a given constant,
find P(θ>π/3).
Let’s try and solve this sub-problem:
P(θ>π/3)=P(tan(θ)>√3)=P(A/B>√3)=P(A>B√3)Now, since A is a Uniform(0,1) random variable, clearly:
P(A>x)=max(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} = dBwhere 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}) dBStep 3: Solving the IntegralPermalink
Expanding the integral:
\int_{0}^{1/\sqrt{3}} (1 - B\sqrt{3}) dBWe 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 .
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