Animated Solution for Mathematics - Probability: Comprehension Passage

Decoding the Urn Problem

Welcome, future engineer. Today, we are not just solving a probability problem; we are peeling back the layers of a system. Probability is the language of uncertainty, and in this problem, we are going to master how to navigate weighted events, conditional logic, and the beautiful convergence of limits.

The Anatomy of the Urn

Imagine you are standing in a room with $n$ urns. Each urn is distinct, labeled $1$ through $n$. The $i^{th}$ urn is a microcosm of the whole system: it contains $n+1$ balls in total.

Within this specific urn, the number of white balls is exactly $i$. This means the probability of drawing a white ball, given that you have chosen the $i^{th}$ urn, is defined by the ratio:

This is our foundational building block. It is a simple, elegant relationship. As the index $i$ increases, the 'whiteness' of the urn increases.

The Proportionality Trap

Now, here is where many students stumble. We are told $P(u_i)\propto i$. This is not a uniform distribution, so we cannot simply say $P(u_i)=1/n$.

We must introduce a proportionality constant, $k$, such that $P(u_i)=ki$. To find $k$, we rely on the fundamental axiom of probability: the sum of all possible outcomes must equal $1$.

We know the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$. Substituting this in, we get $k\cdot \frac{n(n+1)}{2}=1$, which yields our constant:

This constant is the 'weight' of our system. It ensures that the probability distribution is normalized.

The Law of Total Probability

With our weights defined, we can now calculate the total probability of drawing a white ball, $P(w)$. We use the Law of Total Probability, which tells us to sum the conditional probabilities weighted by the probability of choosing each urn:

Substituting our values:

We can pull the constants out of the summation:

This is the moment where the algebra becomes satisfying. We are summing the squares of the first $n$ integers, which is $\frac{n(n+1)(2n+1)}{6}$. Let us plug it in:

Watch the cancellation! The $n$ cancels, one $(n+1)$ cancels, and the $2$ and $6$ simplify. We are left with:

The Asymptotic Limit

The problem asks for the limit as $n\to \infty$. This is where we look at the behavior of the system as it grows infinitely large. We divide the numerator and denominator by $n$:

It is a beautiful result. Despite the complexity of the urns, the system converges to a clean $2/3$.

Bayes' Theorem and Even Urns

In the second part, we assume $P(u_i)=c=1/n$. The calculation simplifies significantly, leading to $P(w)=1/2$. When we apply Bayes' Theorem to find $P(u_n|w)$, we are essentially asking: 'Given that I have a white ball, what is the probability it came from the last urn?'

The math follows directly:

Finally, for the even-numbered urns, we restrict our sample space. We define event $E$ as choosing an even urn, where $P(E)=1/2$. By summing over $j$ where $i=2j$, we find $P(w\cap E)=\frac{n+2}{4(n+1)}$.

Dividing by $P(E)$ gives us the final conditional probability:

You have navigated through proportionality, summation, limits, and Bayes' Theorem. This is the essence of JEE Advanced mathematics—taking a complex, layered problem and breaking it down into manageable, logical steps. Keep this clarity, and you will conquer any problem that comes your way.