Seven white balls and three black balls are randomly placed in a row. The probability that no two bl

Q: Why can't we just use the total arrangements minus the cases where they are together?

That would be a combinatorial nightmare! You would have to account for cases where two are together, three are together, and overlapping sets. The Gap Method is much cleaner because it directly constructs the favorable arrangements without needing to subtract complex cases.

Q: Why are there $8$ gaps and not $7$?

Think of the white balls as separators. If you have $7$ balls, you have $6$ spaces between them, plus one at the very start and one at the very end. That is $6+2=8$ total gaps. If you only counted the spaces between, you would miss the arrangements where a black ball is at the very beginning or end.

Q: Does the order of the white balls matter?

No, the problem states they are identical. If they were distinct, we would multiply by $7!$, but since they are identical, we only care about the positions. The probability remains the same regardless of whether the balls are distinct or identical, as long as the counting method is consistent.

Question

Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals

Options:

(A) $1/2$

(B) $7/15$

(C) $2/15$

(D) $1/3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a row of ten empty slots, holding seven white balls and three black balls. The challenge is to arrange these balls such that no two black balls are ever neighbors.

This is a classic problem in combinatorics, and it serves as the perfect gateway to understanding the **Gap Method**.

### The Total Universe of Possibilities

Before we impose any constraints, let us look at the total number of ways to arrange these ten balls. We have ten positions in total and we need to choose three of them for our black balls.

Since the balls of the same color are identical, the order in which we place them does not matter. This is a straightforward combination problem. The total number of arrangements, denoted as $n(S)$, is given by $\binom{10}{3}$.

Calculating this, we have:

$$n(S) = \binom{10}{3} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$$

So, there are $120$ possible ways to arrange these balls in a row.

### The Elegant Solution: The Gap Method

Now, let us tackle the constraint: no two black balls can be adjacent. Instead of trying to count the 'bad' arrangements and subtracting them, we use the **Gap Method**.

We start by placing the seven identical white balls in a row. This creates a series of gaps. Think of the white balls as separators.

If you have seven white balls, you have a gap before the first ball, a gap between each pair of balls, and a gap after the last ball. Counting these, we find $7+1=8$ available gaps.

If we place at most one black ball in any of these gaps, it is physically impossible for any two black balls to be adjacent. This is the beauty of the method; it turns a complex constraint into a simple selection problem.

### The Final Calculation

We have $8$ gaps and we need to choose $3$ of them to place our $3$ black balls. The number of favorable arrangements, $n(E)$, is $\binom{8}{3}$.

Calculating this, we get:

$$n(E) = \binom{8}{3} = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$$

Now, we have our favorable outcomes ($56$) and our total outcomes ($120$). The probability $P(E)$ is simply the ratio of favorable outcomes to total outcomes:

$$P(E) = \frac{56}{120}$$

To simplify this, we divide both the numerator and the denominator by their greatest common divisor, which is $8$. This gives us:

$$P(E) = \frac{56 \div 8}{120 \div 8} = \frac{7}{15}$$

Through the power of the Gap Method, we have navigated the complexity of the problem and arrived at the elegant result of **$\frac{7}{15}$**. Keep practicing these methods, and you will find that even the most daunting probability problems become a joy to solve.

Animated Solution for Mathematics - Probability: Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently equals

Select Answer:

Visualized Solution

Understanding the Setup

Total Possible Arrangements

Calculating $n (S)$

The Gap Method Strategy

Identifying Available Gaps

Choosing Gaps for Black Balls

Calculating $n (E)$

Finding the Probability

Simplifying the Fraction

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Total Universe of Possibilities

The Elegant Solution

The Final Calculation

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