For three events A, B and C, P (Exactly one of A or B occurs)
= P (Exactly one of B or C occurs)
= P (Exactly one of C or A occurs) = `(1)/(4)` and
P(All the three events occurs simultaneously = `(1)/(16)`.
Then the probability that at least one of the events occurs, is :

Options

A. `(3)/(16)`
B. `(7)/(32)`
C. `(7)/(16)`
D. `(7)/(64)`

Answer

To solve the problem, we need to find the probability that at least one of the events A, B, or C occurs, given the provided probabilities.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We know:
- \( P(\text{Exactly one of } A \text{ or } B) = \frac{1}{4} \)
- \( P(\text{Exactly one of } B \text{ or } C) = \frac{1}{4} \)
- \( P(\text{Exactly one of } C \text{ or } A) = \frac{1}{4} \)
- \( P(A \cap B \cap C) = \frac{1}{16} \)

2. **Using the Formula for Exactly One Event**:
The probability of exactly one of two events occurring can be expressed as:
\[ P(A \text{ or } B) - P(A \cap B) \]
For example, for events A and B:
\[ P(A \text{ or } B) = P(A) + P(B) - P(A \cap B) \]
Therefore, we can write:
\[ P(A \text{ or } B) - P(A \cap B) = \frac{1}{4} \]
This leads us to:
\[ P(A) + P(B) - P(A \cap B) - P(A \cap B) = \frac{1}{4} \]
Simplifying gives:
\[ P(A) + P(B) - 2P(A \cap B) = \frac{1}{4} \quad \text{(1)} \]

3. **Setting Up Similar Equations**:
We can set up similar equations for the other pairs:
- For B and C:
\[ P(B) + P(C) - 2P(B \cap C) = \frac{1}{4} \quad \text{(2)} \]
- For C and A:
\[ P(C) + P(A) - 2P(C \cap A) = \frac{1}{4} \quad \text{(3)} \]

4. **Summing the Equations**:
Adding equations (1), (2), and (3):
\[ (P(A) + P(B) - 2P(A \cap B)) + (P(B) + P(C) - 2P(B \cap C)) + (P(C) + P(A) - 2P(C \cap A)) = \frac{3}{4} \]
This simplifies to:
\[ 2(P(A) + P(B) + P(C)) - 2(P(A \cap B) + P(B \cap C) + P(C \cap A)) = \frac{3}{4} \]
Dividing by 2:
\[ P(A) + P(B) + P(C) - (P(A \cap B) + P(B \cap C) + P(C \cap A)) = \frac{3}{8} \quad \text{(4)} \]

5. **Finding the Probability of At Least One Event**:
The probability of at least one of the events occurring can be found using:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - (P(A \cap B) + P(B \cap C) + P(C \cap A)) + P(A \cap B \cap C) \]
Substituting from equation (4) and the value of \( P(A \cap B \cap C) \):
\[ P(A \cup B \cup C) = \frac{3}{8} + \frac{1}{16} \]
To add these fractions, we need a common denominator:
\[ \frac{3}{8} = \frac{6}{16} \]
Thus:
\[ P(A \cup B \cup C) = \frac{6}{16} + \frac{1}{16} = \frac{7}{16} \]

### Final Answer:
The probability that at least one of the events occurs is \( \frac{7}{16} \).