A bag contains 10 balls out of which k are red and (10 - k) are black, where 0 ≤ k ≤ 10. If three ba

Q: Wait, why do we assume $P(k) = 1/11$? Couldn't the distribution be different?

That is a brilliant question. In probability theory, when we have no information about the distribution of a parameter (in this case, $k$), we invoke the Principle of Indifference. This principle states that if there is no evidence to favor one outcome over another, we must assign them equal probabilities. Since $k$ can be any integer from $0$ to $10$, there are $11$ possible values, hence $P(k) = 1/11$.

Q: Why can't we just use simple probability instead of the Law of Total Probability?

Simple probability works when the composition of the bag is fixed and known. Here, the composition is a variable! We are essentially dealing with a 'mixture' of scenarios. The Law of Total Probability is the mathematical bridge that allows us to account for all possible 'versions' of the bag, weighted by how likely each version is to exist.

Q: Is there a trick to the summation $\sum_{i=3}^{10} \binom{i}{3}$?

Yes, and it is a classic JEE Advanced favorite! It is called the Hockey-stick identity. It states that the sum of binomial coefficients $\binom{r}{r} + \binom{r+1}{r} + \dots + \binom{n}{r}$ is equal to $\binom{n+1}{r+1}$. It is called the 'hockey-stick' because if you visualize Pascal's triangle, the terms form a diagonal line, and the sum is the term 'hooked' below the last element. It turns a tedious addition into a single, elegant combination.

Question

A bag contains 10 balls out of which $k$ are red and $(10 - k)$ are black, where $0 \le k \le 10$. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

Options:

(A) $\frac{14}{55}$

(B) $\frac{7}{55}$

(C) $\frac{7}{110}$

(D) $\frac{7}{11}$

Sigma Percentile · Accepted Answer

### The Mystery of the Bag: A Journey into Bayesian Probability

Imagine you are standing before a bag containing exactly ten balls. Some are red, some are black. You are told there are $k$ red balls and $10-k$ black balls, but the value of $k$ is a complete mystery. It could be anything from zero to ten.

In the world of JEE Advanced, we often encounter problems where the 'state of the world' is uncertain, and we must use the evidence we observe to infer the truth. This is the essence of Bayesian inference.

### Phase 1: The Principle of Indifference

Before we draw any balls, what do we know? We know $k \in \{0, 1, 2, \dots, 10\}$.

Since the problem provides no information about how the bag was filled, we must assume that every possible value of $k$ is equally likely. With eleven possible values, the prior probability for any specific $k$ is:

$$P(k) = \frac{1}{11}$$

This is our starting assumption, our 'prior' belief before we see any data.

### Phase 2: The Observed Reality

Now, we reach into the bag and draw three balls at random without replacement. The result is striking: all three are black. Let us call this Event $E$.

This observation is our data. We need to calculate the likelihood of this event occurring for any given $k$. If there are $k$ red balls, there are $10-k$ black balls. The probability of drawing three black balls is the number of ways to choose three black balls divided by the total number of ways to choose three balls from ten:

$$P(E|k) = \frac{\binom{10-k}{3}}{\binom{10}{3}}$$

This formula is our likelihood function.

### Phase 3: The Law of Total Probability

To understand the probability of observing three black balls across all possible bags, we use the Law of Total Probability. We sum the likelihood of $E$ for each $k$, weighted by the prior probability $P(k)$:

$$P(E) = \sum_{k=0}^{10} P(E|k) \cdot P(k) = \sum_{k=0}^{10} \left( \frac{\binom{10-k}{3}}{\binom{10}{3}} \cdot \frac{1}{11} ight)$$

At first glance, this summation looks daunting. But let us simplify. We pull the constants $\frac{1}{11 \cdot \binom{10}{3}}$ outside the sum. We are left with $\sum_{k=0}^{10} \binom{10-k}{3}$.

If we substitute $i = 10-k$, as $k$ goes from $0$ to $10$, $i$ goes from $10$ down to $0$. The sum becomes $\sum_{i=3}^{10} \binom{i}{3}$.

This is where the magic of the **Hockey-stick identity** happens. The identity $\sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1}$ allows us to collapse this entire sum into $\binom{11}{4}$.

Calculating these values, $\binom{11}{4} = 330$ and $\binom{10}{3} = 120$. Thus:

$$P(E) = \frac{1}{11 \cdot 120} 	imes 330 = \frac{330}{1320} = \frac{1}{4}$$

The total probability of drawing three black balls is exactly one-fourth.

### Phase 4: Bayes' Theorem

We have the prior, the likelihood, and the total probability. Now, we use Bayes' Theorem to find the posterior probability: the probability that the bag contains exactly one red ball ($k=1$), given that we observed three black balls.

The formula is:

$$P(k=1|E) = \frac{P(E|k=1) \cdot P(k=1)}{P(E)}$$

For $k=1$, the likelihood is:

$$P(E|k=1) = \frac{\binom{9}{3}}{\binom{10}{3}} = \frac{84}{120} = \frac{7}{10}$$

Plugging this into Bayes' formula:

$$P(k=1|E) = \frac{\frac{7}{10} \cdot \frac{1}{11}}{1/4} = \frac{7/110}{1/4} = \frac{7}{110} 	imes 4 = \frac{28}{110} = \frac{14}{55}$$

And there we have it! The probability is **$\frac{14}{55}$**. This problem is a beautiful demonstration of how we can use evidence to update our beliefs about an uncertain world.

Select Answer:

Visualized Solution

Understanding the Scenario

Defining Prior Probabilities $P (k)$

The Observed Event $E$

Likelihood of Drawing 3 Black Balls

Total Probability $P (E)$ Setup

Evaluating the Summation

Applying Hockey-stick Identity

Calculating Total Probability $P (E)$

The Specific Case: $k = 1$

Applying Bayes' Theorem

Final Calculation

The Sigma Insight: Bayes' Theorem

The Mystery of the Bag

Phase 1

Phase 2

Phase 3

Phase 4

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