Of three independent events, the chance that only the first occurs is a, the chance that only the second occurs is b and the chance of only third is c. if x is a root of the equation `(a + x) (b + x) ( c + x) = x^(2)` , then chance of three events are respectively :

Options

A. `a // (a + x ) , b// (b + x ) , c// (c + x)`
B. `x // (a + x ) , x// ( b + x ) , x// (c + x)`
C. `a // (a + 2 x) , b// (b + 2x) , c // ( c + 2x)`
D. `2x // (a + 2x) , 2x// (b + 2x) , 2x// (c + 2x)`

Answer

To solve the problem, we need to find the probabilities of three independent events E1, E2, and E3, given the chances that only the first, second, and third events occur are a, b, and c respectively. We are also given that x is a root of the equation \((a + x)(b + x)(c + x) = x^2\).

### Step-by-Step Solution:

1. **Understanding the Events**:
- Let \( P(E1) \) be the probability of event E1 occurring, \( P(E2) \) for event E2, and \( P(E3) \) for event E3.
- The probabilities of the complements are \( P(E1^c) \), \( P(E2^c) \), and \( P(E3^c) \).

2. **Setting Up the Probabilities**:
- The probability that only E1 occurs is given by:
\[ P(E1) \cdot P(E2^c) \cdot P(E3^c) = a \]
- The probability that only E2 occurs is:
\[ P(E1^c) \cdot P(E2) \cdot P(E3^c) = b \]
- The probability that only E3 occurs is:
\[ P(E1^c) \cdot P(E2^c) \cdot P(E3) = c \]

3. **Expressing the Probabilities**:
- Let \( P(E1) = p_1 \), \( P(E2) = p_2 \), and \( P(E3) = p_3 \).
- Then, we can express the equations as:
\[ p_1 (1 - p_2)(1 - p_3) = a \quad (1) \]
\[ (1 - p_1) p_2 (1 - p_3) = b \quad (2) \]
\[ (1 - p_1)(1 - p_2) p_3 = c \quad (3) \]

4. **Dividing the Equations**:
- From equation (1) divided by (2):
\[ \frac{p_1}{(1 - p_1)} = \frac{a}{b} \]
Hence,
\[ p_1 = \frac{a}{a + b} \]
- From equation (2) divided by (3):
\[ \frac{p_2}{(1 - p_2)} = \frac{b}{c} \]
Hence,
\[ p_2 = \frac{b}{b + c} \]
- From equation (3) divided by (1):
\[ \frac{p_3}{(1 - p_3)} = \frac{c}{a} \]
Hence,
\[ p_3 = \frac{c}{c + a} \]

5. **Final Probabilities**:
- Therefore, the probabilities of the events are:
\[ P(E1) = \frac{a}{a + b}, \quad P(E2) = \frac{b}{b + c}, \quad P(E3) = \frac{c}{c + a} \]

### Conclusion:
The chances of the three events E1, E2, and E3 occurring are respectively:
- \( P(E1) = \frac{a}{a + b} \)
- \( P(E2) = \frac{b}{b + c} \)
- \( P(E3) = \frac{c}{c + a} \)