If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the bo

Q: Wait, the question says identical balls. Why are we using combinations as if they are distinct?

Great question! In probability, we need a sample space where every outcome has the same chance of occurring. If we treat balls as identical, the outcome 'all balls in box 1' is much less likely than 'balls spread out'. By treating them as distinct, we ensure each ball has an independent choice, making every outcome equally likely.

Q: Why $3^{12}$? Shouldn't it be stars and bars?

Stars and bars is for counting *ways* to distribute identical items, but it doesn't give you *equally likely* outcomes for probability. For probability, we need the total number of ways each ball can choose a box, which is $3 \times 3 \times ...$ (12 times).

Q: Does 'one of the boxes' mean *exactly* one or *at least* one?

In this context, it implies a specific box (like 'Box A') contains exactly 3 balls. We calculate the probability for that specific box and the math naturally handles the rest.

Question

If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is

Options:

(A) $220(1/3)^{12}$

(B) $22(1/3)^{11}$

(C) $\frac{55}{3}(2/3)^{11}$

(D) $55(2/3)^{10}$

Sigma Percentile · Accepted Answer

### Analyzing the Probability Paradox

The problem states that we have twelve balls and three boxes. While the balls and boxes are described as identical, we must treat them as **distinct** to ensure that every outcome in our sample space is equally likely.

If we were to treat them as identical, we would be counting arrangements rather than probabilities, which would lead to an incorrect distribution of likelihoods. By labeling the balls $B_1, B_2, ..., B_{12}$, we ensure that each ball makes an independent choice, creating a uniform sample space.

### The Sample Space: The Power of Choice

To build our sample space, we consider that each of the $12$ balls has $3$ choices of boxes. Since each ball's choice is independent, we multiply the possibilities for each ball.

The total number of equally likely outcomes is:

$$n(S) = 3 	imes 3 	imes \dots 	imes 3 	ext{ (12 times)} = 3^{12}$$

This value serves as the denominator for our probability calculation.

### The Favorable Event: Focusing the Lens

The question asks for the probability that a specific box contains exactly $3$ balls. Let us fix our attention on Box $1$.

First, we choose which $3$ balls out of the $12$ will reside in Box $1$. The number of ways to select these balls is given by the combination formula:

$${}^{12}C_3 = \frac{12!}{3!(12-3)!}$$

Next, we have $9$ balls remaining. These $9$ balls must be distributed among the remaining $2$ boxes, and each of these balls has $2$ choices. Therefore, the number of ways to distribute the remaining balls is $2^9$.

The total number of favorable outcomes for this specific box is:

$$n(E) = {}^{12}C_3 	imes 2^9$$

### The Final Calculation: Elegant Simplification

Now, we assemble our probability $P(E) = \frac{n(E)}{n(S)}$. Substituting our values, we get:

$$P(E) = \frac{{}^{12}C_3 	imes 2^9}{3^{12}}$$

We calculate the combination as follows:

$${}^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$$

We can express $220$ as $55 	imes 4$, or $55 	imes 2^2$. Substituting this back into our equation:

$$P(E) = \frac{55 	imes 2^2 	imes 2^9}{3^{12}} = \frac{55 	imes 2^{11}}{3^{12}}$$

To reach the final form, we separate one $3$ from the denominator:

$$P(E) = \frac{55}{3} 	imes \frac{2^{11}}{3^{11}} = \mathbf{\frac{55}{3} \left(\frac{2}{3}ight)^{11}}$$

This result demonstrates the importance of distinguishing between counting arrangements and calculating probabilities in a uniform sample space.

Animated Solution for Mathematics - Probability: If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is

Select Answer:

Visualized Solution

The Probability Paradox

Total Sample Space

Favorable Event Setup

Distributing Remaining Balls

The Probability Fraction

Simplifying $^{12} C_{3}$

Final Answer

The Sigma Insight: Classical Definition of Probability

Analyzing the Probability Paradox

The Sample Space

The Favorable Event

The Final Calculation

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