A box 'A' contains 2 white, 3 red and 2 black balls. Another box 'B' contains 4 white, 2 red and 3 b

Q: Wait, why do we use combinations $\binom{n}{r}$ here? Can't I just multiply probabilities?

Great question! You *could* use simple probability multiplication, but you must be careful about the order. If you pick a White ball then a Red ball, that's one sequence. If you pick Red then White, that's another. Using combinations $\binom{n}{r}$ automatically accounts for all possible orderings of the selection, saving you from double-counting or missing cases. It is the safer, more robust path.

Q: Why can we just cancel out the $\frac{1}{2}$ in the Bayes' formula?

Mathematically, it's because $\frac{1}{2}$ appears in every term of the numerator and the denominator. Conceptually, it means that since both boxes are equally likely to be chosen initially, the 'prior' probability doesn't bias our result. We only care about the 'likelihood' of the event occurring within each box.

Q: What if the balls were drawn *with* replacement? Would the answer change?

Absolutely! If we drew with replacement, the total number of balls would remain constant for the second draw, and the probability of picking a specific color would be independent for each draw. The denominator would change from $\binom{n}{2}$ to $n \times n$. Always check the 'without replacement' clause—it's a classic JEE trap!

Question

A box 'A' contains 2 white, 3 red and 2 black balls. Another box 'B' contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box 'B' is :

Options:

(A) 7/16

(B) 7/8

(C) 9/16

(D) 9/32

Sigma Percentile · Accepted Answer

### The Detective's Dilemma: Unraveling Bayes' Theorem

Welcome, future engineer. Today, we are not just solving a probability problem; we are stepping into the shoes of a detective.

Imagine you are presented with two sealed boxes, Box A and Box B. You reach in, pull out two balls, and see one white and one red.

The question is simple, yet profound: Given this evidence, what is the probability that you were holding Box B? This is the essence of inverse probability, and it is the domain of the legendary Bayes' Theorem.

### Phase 1: Setting the Stage

Before we dive into the algebra, let us visualize the reality of our boxes. Box A is a collection of $2$ white, $3$ red, and $2$ black balls, totaling $7$ balls.

Box B is slightly larger, holding $4$ white, $2$ red, and $3$ black balls, totaling $9$ balls.

We define our event $E$ as the act of drawing exactly one white ball and one red ball. Our goal is to find the conditional probability $P(B|E)$, which reads as 'the probability of Box B given that event $E$ has occurred.'

### Phase 2: The Philosophy of Bayes' Theorem

Many students fear Bayes' Theorem, but it is actually quite intuitive. It is a bridge between the 'prior' probability (what we know before the experiment) and the 'posterior' probability (what we know after seeing the evidence).

The formula is:

$$P(B|E) = \frac{P(B) \cdot P(E|B)}{P(A) \cdot P(E|A) + P(B) \cdot P(E|B)}$$

Since we choose a box at random, $P(A) = P(B) = \frac{1}{2}$. This is our 'prior.' Now, we need the 'likelihoods'—the probability of drawing our specific pair of balls from each box individually.

### Phase 3: The Investigation of Box A

Let us peer into Box A. We need to choose $1$ white ball from $2$ and $1$ red ball from $3$. The total number of ways to choose any $2$ balls from $7$ is $\binom{7}{2}$.

Using the combination formula, we calculate:

$$P(E|A) = \frac{\binom{2}{1} 	imes \binom{3}{1}}{\binom{7}{2}} = \frac{2 	imes 3}{21} = \frac{6}{21} = \frac{2}{7}$$

This is the probability that, if we were holding Box A, we would see the evidence we currently have. It is a crucial piece of our puzzle.

### Phase 4: The Investigation of Box B

Now, we turn our attention to Box B. The logic remains identical, but the numbers shift. We need $1$ white ball from $4$ and $1$ red ball from $2$.

The total ways to draw $2$ balls from $9$ is $\binom{9}{2}$.

$$P(E|B) = \frac{\binom{4}{1} 	imes \binom{2}{1}}{\binom{9}{2}} = \frac{4 	imes 2}{36} = \frac{8}{36} = \frac{2}{9}$$

Notice how the likelihood of our event $E$ changes depending on the box. This difference is exactly what allows us to 'reverse' the probability.

### Phase 5: The Grand Synthesis

Now, we bring it all together. We substitute our values into the Bayes' formula.

Because $P(A)$ and $P(B)$ are both $\frac{1}{2}$, they appear in every term of the fraction. We can elegantly cancel them out, simplifying our expression significantly:

$$P(B|E) = \frac{\frac{1}{2} \cdot \frac{2}{9}}{\frac{1}{2} \cdot \frac{2}{7} + \frac{1}{2} \cdot \frac{2}{9}} = \frac{\frac{2}{9}}{\frac{2}{7} + \frac{2}{9}}$$

We can further simplify by dividing everything by $2$:

$$P(B|E) = \frac{\frac{1}{9}}{\frac{1}{7} + \frac{1}{9}}$$

Now, for the final arithmetic. We find a common denominator for the bottom fraction, which is $63$:

$$P(B|E) = \frac{\frac{1}{9}}{\frac{9+7}{63}} = \frac{1}{9} 	imes \frac{63}{16}$$

After canceling the $9$ into the $63$, we arrive at our final result: **$\frac{7}{16}$**.

### Conclusion

And there it is. The probability that our balls originated from Box B is **$\frac{7}{16}$**. It is a beautiful, clean result.

Remember, in JEE Advanced, the math is rarely the hardest part—it is the clarity of thought. By breaking the problem into the 'prior' and the 'likelihood,' you turn a complex probability question into a simple, logical sequence. Keep practicing this mindset, and you will master the art of the problem-solver.

Select Answer:

Visualized Solution

Understanding the Setup

Applying Bayes' Theorem

Calculating $P (E ∣ A)$

Calculating $P (E ∣ B)$

Substitution in Bayes' Formula

Simplifying the Expression

Final Calculation

Conclusion

The Sigma Insight: Bayes' Theorem

The Detective's Dilemma

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Conclusion

Similar Questions

Comprehension Passage

If 1 ball is drawn from each of the boxes $B_{1}, B_{2}$ and $B_{3}$ , the probability that all 3 drawn balls are of the same colour is

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Comprehension Passage

A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 1/3, then the correct option(s) with the possible values of $n_{1}$ and $n_{2}$ is(are)

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Comprehension Passage

The probability of the drawn ball from $U_{2}$ being white is

Given that the drawn ball from $U_{2}$ is white, the probability that head appeared on the coin is

Select Answer:

Visualized Solution

Understanding the Setup

Applying Bayes' Theorem

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣A)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣B)

Substitution in Bayes' Formula

Simplifying the Expression

Final Calculation

Conclusion

The Sigma Insight: Bayes' Theorem

The Detective's Dilemma

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Conclusion

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1 ball is drawn from each of the boxes B1​,B2​ and B3​, the probability that all 3 drawn balls are of the same colour is

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Comprehension Passage

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The probability of the drawn ball from U2​ being white is

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Calculating $P (E ∣ A)$

Calculating $P (E ∣ B)$

If 1 ball is drawn from each of the boxes $B_{1}, B_{2}$ and $B_{3}$ , the probability that all 3 drawn balls are of the same colour is

The probability of the drawn ball from $U_{2}$ being white is

Given that the drawn ball from $U_{2}$ is white, the probability that head appeared on the coin is