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 occur simultaneously) `=1/16dot` Then the probability that at least one of the events occurs, is : `7/(64)` (2) `3/(16)` (3) `7/(32)` (4) `7/(16)`

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the probabilities
Let:
- \( P(A) = p_A \)
- \( P(B) = p_B \)
- \( P(C) = p_C \)
- \( P(A \cap B) = p_{AB} \)
...