If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 b

Q: Wait, why do we multiply the probabilities for each case?

Great question! We multiply because the draws from each box are independent events. The outcome of Box 1 does not influence the outcome of Box 2 or Box 3. In probability theory, for independent events $A$, $B$, and $C$, the probability of all three occurring is $P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C)$.

Q: Why do we add the probabilities of the three cases together?

We add them because these cases are mutually exclusive. You cannot simultaneously have the black ball in Box 1 AND in Box 2. Since only one of these scenarios can happen at a time to satisfy the condition of 'exactly 2 white and 1 black', we sum their individual probabilities to find the total probability of the event.

Q: What if the boxes had different total numbers of balls? Would the logic change?

The fundamental logic remains the same! You would still identify the mutually exclusive cases and multiply the probabilities within each case. The only difference would be that the denominator for each probability would depend on the total number of balls in that specific box. Always focus on the structure of the event first!

Question

If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is

Options:

(A) $13/32$

(B) $1/4$

(C) $1/32$

(D) $3/16$

Sigma Percentile · Accepted Answer

### The Anatomy of the Problem

Imagine you are standing before three distinct boxes. Each box is a small universe of its own, containing a specific mix of white and black balls.

Box 1 holds $3$ white and $1$ black ball. Box 2 is a perfect $2$ white and $2$ black split. Box 3 is the inverse of Box 1, with $1$ white and $3$ black balls.

You are tasked with drawing exactly one ball from each box. Your goal is to hold exactly two white balls and one black ball in your hands.

### The Logic of Cases

To achieve our goal of $2$ White and $1$ Black, we must realize that the single black ball can only originate from one of the three boxes. This realization is the key to unlocking the problem.

We can break this down into three mutually exclusive scenarios:

*   **Case 1:** We draw a White ball from Box 1, a White ball from Box 2, and a Black ball from Box 3.
*   **Case 2:** We draw a White ball from Box 1, a Black ball from Box 2, and a White ball from Box 3.
*   **Case 3:** We draw a Black ball from Box 1, a White ball from Box 2, and a White ball from Box 3.

Because these cases are mutually exclusive, we will calculate the probability of each and then sum them up at the end.

### The Calculation

Since the draws are independent, we multiply the probabilities of the individual events for each case.

For **Case 1** ($W_1, W_2, B_3$):

$$P(	ext{Case 1}) = P(W_1) \cdot P(W_2) \cdot P(B_3) = \frac{3}{4} \cdot \frac{2}{4} \cdot \frac{3}{4} = \frac{18}{64}$$

For **Case 2** ($W_1, B_2, W_3$):

$$P(	ext{Case 2}) = P(W_1) \cdot P(B_2) \cdot P(W_3) = \frac{3}{4} \cdot \frac{2}{4} \cdot \frac{1}{4} = \frac{6}{64}$$

For **Case 3** ($B_1, W_2, W_3$):

$$P(	ext{Case 3}) = P(B_1) \cdot P(W_2) \cdot P(W_3) = \frac{1}{4} \cdot \frac{2}{4} \cdot \frac{1}{4} = \frac{2}{64}$$

### The Synthesis

Now, we bring it all together. The total probability $P(2W, 1B)$ is the sum of these three cases:

$$P(2W, 1B) = \frac{18}{64} + \frac{6}{64} + \frac{2}{64} = \frac{26}{64}$$

Finally, we simplify the fraction. Dividing both the numerator and the denominator by $2$, we get **$\frac{13}{32}$**.

This is our final answer, a beautiful result derived from a systematic approach to a seemingly complex problem. In JEE Advanced, the secret is not just in the calculation, but in the clarity of your case-based thinking.

Animated Solution for Mathematics - Probability: If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is

Select Answer:

Visualized Solution

Visualizing the Three Boxes

Identifying the Three Cases

Case 1: $W_{1}, W_{2}, B_{3}$

Case 2: $W_{1}, B_{2}, W_{3}$

Case 3: $B_{1}, W_{2}, W_{3}$

Summing the Probabilities

Simplifying the Fraction

The Way Forward

The Sigma Insight: Addition and Multiplication Theorems

The Anatomy of the Problem

The Logic of Cases

The Calculation

The Synthesis

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