Seven white balls and three black balls are randomly placed in a row. What is the probability that no two black balls are placed adjecently ?

Options

A. `(7)/(15)`
B. `(8)/(15)`
C. `(11)/(15)`
D. `(13)/(15)`

Answer

To solve the problem of finding the probability that no two black balls are placed adjacently when 7 white balls and 3 black balls are randomly arranged in a row, we can follow these steps:

### Step 1: Calculate the Total Arrangements of Balls
First, we need to find the total number of arrangements of 10 balls (7 white and 3 black). The formula for the total arrangements of n objects where there are groups of identical objects is given by:

\[ N(S) = \frac{n!}{n_1! \cdot n_2!} \]

In our case, we have 10 balls in total (7 white and 3 black):

\[ N(S) = \frac{10!}{7! \cdot 3!} \]

### Step 2: Calculate the Arrangements with No Two Black Balls Together
Next, we need to arrange the balls such that no two black balls are adjacent. To do this, we first arrange the 7 white balls in a row. This creates 8 possible gaps (including the ends) where the black balls can be placed:

- _W_ _W_ _W_ _W_ _W_ _W_ _W_

The underscores represent the gaps where the black balls can be placed.

### Step 3: Choose Gaps for Black Balls
We need to choose 3 out of these 8 gaps to place the black balls. The number of ways to choose 3 gaps from 8 is given by the combination formula:

\[ N(A) = \binom{8}{3} \]

### Step 4: Calculate the Probability
Now, we can find the probability that no two black balls are adjacent. The probability \( P(A) \) is given by the ratio of the number of favorable outcomes to the total outcomes:

\[ P(A) = \frac{N(A)}{N(S)} \]

Substituting the values we calculated:

\[ P(A) = \frac{\binom{8}{3}}{\frac{10!}{7! \cdot 3!}} \]

### Step 5: Simplify the Probability Expression
Now we simplify the expressions:

1. Calculate \( N(S) \):
\[ N(S) = \frac{10!}{7! \cdot 3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

2. Calculate \( N(A) \):
\[ N(A) = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

3. Now substitute back into the probability:
\[ P(A) = \frac{56}{120} = \frac{14}{30} = \frac{7}{15} \]

### Final Answer
The probability that no two black balls are placed adjacently is:

\[ \frac{7}{15} \]