Seven white balls and three black balls are randomly placed in a row. IF the parabability that no two black balls are placed adjacently, equals `(7/a),` then the value of "a" is

Answer

To solve the problem of finding the value of "a" such that the probability of placing 7 white balls and 3 black balls in a row with no two black balls adjacent equals \( \frac{7}{a} \), we will follow these steps:

### Step-by-Step Solution:

1. **Total Arrangements**:
We first calculate the total number of arrangements of 7 white balls and 3 black balls. The total number of ways to arrange these 10 balls (7 white and 3 black) is given by the combination formula:
\[ \text{Total arrangements} = \binom{10}{3} \]
This represents choosing 3 positions out of 10 for the black balls.

2. **Arrangements with No Adjacent Black Balls**:
To ensure that no two black balls are adjacent, we can first place the 7 white balls in a row. This creates 8 gaps (including the ends) where the black balls can be placed:
- Gaps: _ W _ W _ W _ W _ W _ W _ W _
Thus, we have 8 gaps to place the 3 black balls.

3. **Choosing Gaps for Black Balls**:
We need to choose 3 out of these 8 gaps to place the black balls. The number of ways to do this is:
\[ \text{Favorable arrangements} = \binom{8}{3} \]

4. **Calculating the Probability**:
The probability that no two black balls are adjacent is given by the ratio of the favorable arrangements to the total arrangements:
\[ P(\text{no two black balls adjacent}) = \frac{\binom{8}{3}}{\binom{10}{3}} \]

5. **Calculating Combinations**:
Now, we calculate the combinations:
\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]
\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

6. **Substituting Values into Probability**:
Now substituting these values back into the probability formula:
\[ P = \frac{56}{120} = \frac{14}{30} = \frac{7}{15} \]

7. **Setting Up the Equation**:
According to the problem, this probability is equal to \( \frac{7}{a} \):
\[ \frac{7}{15} = \frac{7}{a} \]

8. **Solving for "a"**:
By cross-multiplying, we find:
\[ 7a = 7 \times 15 \implies a = 15 \]

### Final Answer:
Thus, the value of \( a \) is \( \boxed{15} \).