Bag B_1 contains 6 white and 4 blue balls, Bag B_2 contains 4 white and 6 blue balls, and Bag B_3 co

Q: Wait, why do we multiply the probabilities? Can't we just look at the number of white balls in each bag?

Great question! If you only look at the number of white balls, you are ignoring the fact that the bags themselves were chosen randomly. Multiplying by $P(B_i) = 1/3$ accounts for the 'weight' of each bag in the selection process. Without it, you are assuming every bag has an equal chance of being the source, which is true, but you must mathematically formalize that choice.

Q: Why does the $1/3$ cancel out in the final step? Is that a coincidence?

It is not a coincidence at all! Because each bag has an equal prior probability of being selected, the $1/3$ appears in every term of the numerator and the denominator. It effectively scales the entire equation, allowing us to focus purely on the relative ratios of the white balls in each bag. It is a beautiful simplification that happens whenever the priors are uniform.

Q: I keep getting confused between $P(W|B_2)$ and $P(B_2|W)$. How do I keep them straight?

Think of the vertical bar $|$ as 'given that'. $P(W|B_2)$ means 'the probability of drawing a White ball, given that we are already looking inside Bag 2'. $P(B_2|W)$ means 'the probability that we are looking at Bag 2, given that we have already seen a White ball'. One is a forward prediction, the other is a backward inference.

Q: Do I really need to write out the full Bayes' Theorem formula every time?

For JEE Advanced, yes! While you might see the shortcut, writing the full formula ensures you don't miss a term in the denominator. It is the difference between a lucky guess and a rigorous, bulletproof derivation that earns you full marks.

Q: What if the bags didn't have the same number of balls? Would the method change?

The logic remains identical! The only thing that would change is the conditional probability $P(W|B_i)$. You would simply calculate the ratio of white balls to total balls in each specific bag. The structure of Bayes' Theorem is universal; it doesn't care about the contents, only the probabilities.

Question

Bag $B_{1}$ contains 6 white and 4 blue balls, Bag $B_{2}$ contains 4 white and 6 blue balls, and Bag $B_3$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag $B_{2}$ is:

Options:

(A) $\frac{4}{15}$

(B) $\frac{1}{3}$

(C) $\frac{2}{5}$

(D) $\frac{2}{3}$

Sigma Percentile · Accepted Answer

### The Mystery of the Hidden Bag: A Journey into Bayes' Theorem

Imagine you are standing in a quiet room. Before you sit three identical, opaque bags, labeled $B_1$, $B_2$, and $B_3$. You know their contents, but you cannot see inside.

You reach out, pick one bag at random, and pull out a single ball. It is white. Now, here is the challenge: can you determine the probability that this white ball originated from Bag $B_2$?

This is not just a probability problem; it is a detective story. We are working backward from an observed effect to identify the hidden cause.

### Phase 1: Setting the Stage

First, let us inventory our evidence. We have three bags with distinct compositions:

- Bag $B_1$: 6 white, 4 blue balls.
- Bag $B_2$: 4 white, 6 blue balls.
- Bag $B_3$: 5 white, 5 blue balls.

Since we choose a bag at random, the prior probability of selecting any specific bag is equal:

$$P(B_1) = P(B_2) = P(B_3) = \frac{1}{3}$$

This is our starting point. We are in a state of perfect uncertainty regarding which bag we hold.

### Phase 2: The Conditional Reality

Now, we consider the 'effect'—the white ball. We must calculate the probability of drawing a white ball from each bag individually. These are our conditional probabilities:

- $P(W|B_1) = \frac{6}{10}$
- $P(W|B_2) = \frac{4}{10}$
- $P(W|B_3) = \frac{5}{10}$

Notice how the probability of drawing a white ball changes depending on which bag we are holding. This is the heart of the problem. We have observed the event $W$, and we want to find the probability of the specific cause $B_2$. Mathematically, we are seeking $P(B_2|W)$.

### Phase 3: The Bridge of Bayes

To solve this, we invoke the power of Bayes' Theorem. It is the ultimate tool for updating our beliefs based on new evidence. The formula is elegant and powerful:

$$P(B_2|W) = \frac{P(B_2)P(W|B_2)}{P(B_1)P(W|B_1) + P(B_2)P(W|B_2) + P(B_3)P(W|B_3)}$$

Look at the structure. The numerator represents the specific path we are interested in: choosing Bag $B_2$ AND drawing a white ball.

The denominator is the sum of all possible paths that could have resulted in a white ball. We are essentially asking: 'Of all the ways to get a white ball, what fraction of them come from Bag $B_2$?'

### Phase 4: The Elegant Cancellation

Now, let us substitute our values. The numerator becomes $\frac{1}{3} 	imes \frac{4}{10}$. The denominator is the sum of the three paths:

$$	ext{Denominator} = \frac{1}{3} 	imes \frac{6}{10} + \frac{1}{3} 	imes \frac{4}{10} + \frac{1}{3} 	imes \frac{5}{10}$$

Here is where the beauty of mathematics reveals itself. Notice that every single term contains a factor of $\frac{1}{3}$ and a factor of $\frac{1}{10}$. We can factor these out:

$$P(B_2|W) = \frac{\frac{1}{3} 	imes \frac{4}{10}}{\frac{1}{3} 	imes \frac{1}{10} (6 + 4 + 5)}$$

Everything cancels out! The $\frac{1}{3}$ and the $\frac{1}{10}$ vanish, leaving us with a simple ratio of the white ball counts:

$$P(B_2|W) = \frac{4}{6 + 4 + 5} = \frac{4}{15}$$

### Conclusion

And there it is: **$\frac{4}{15}$**. We have successfully navigated the uncertainty.

By using Bayes' Theorem, we transformed a confusing scenario into a clear, logical path. Remember, in JEE Advanced, the math is not just about calculation; it is about understanding the flow of information.

When you see a problem like this, do not panic. Define your events, identify your priors, and let the theorem guide you to the truth. You have the tools—now go out and solve the next mystery!

Select Answer:

Visualized Solution

Visualizing the Bags

Defining the Event

Prior Probabilities

Conditional Probabilities

The Goal

Bayes' Theorem Setup

Substituting Values (Numerator)

Substituting Values (Denominator)

Simplification

Final Calculation

The Sigma Insight: Bayes' Theorem

The Mystery of the Hidden Bag

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

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Comprehension Passage

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