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 :

Options

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

Explanation

To solve the problem step by step, we start by defining the probabilities and using the information given in the question.

### Step 1: Define the probabilities
Let:
- \( P(A) = p_a \)
- \( P(B) = p_b \)
- \( P(C) = p_c \)
- \( P(A \cap B) = p_{ab} \)
- \( P(B \cap C) = p_{bc} \)
- \( P(C \cap A) = p_{ca} \)
- \( P(A \cap B \cap C) = p_{abc} = \frac{1}{16} \)

### Step 2: Set up the equations
From the problem statement, we know:
1. \( P(\text{Exactly one of } A \text{ or } B) = p_a + p_b - 2p_{ab} = \frac{1}{4} \)
2. \( P(\text{Exactly one of } B \text{ or } C) = p_b + p_c - 2p_{bc} = \frac{1}{4} \)
3. \( P(\text{Exactly one of } C \text{ or } A) = p_c + p_a - 2p_{ca} = \frac{1}{4} \)

### Step 3: Add the equations
Adding all three equations gives us:
\[ (p_a + p_b - 2p_{ab}) + (p_b + p_c - 2p_{bc}) + (p_c + p_a - 2p_{ca}) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} \]
This simplifies to:
\[ 2p_a + 2p_b + 2p_c - 2(p_{ab} + p_{bc} + p_{ca}) = \frac{3}{4} \]

### Step 4: Divide by 2
Dividing the entire equation by 2 gives:
\[ p_a + p_b + p_c - (p_{ab} + p_{bc} + p_{ca}) = \frac{3}{8} \]

### Step 5: Express the probability of at least one event
The probability of at least one of the events \( A, B, C \) occurring is given by:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C) \]
Substituting the values we have:
\[ P(A \cup B \cup C) = (p_a + p_b + p_c) - (p_{ab} + p_{bc} + p_{ca}) + \frac{1}{16} \]
Using the equation from Step 4:
\[ P(A \cup B \cup C) = \frac{3}{8} + \frac{1}{16} \]

### Step 6: Find a common denominator and add
To add \( \frac{3}{8} \) and \( \frac{1}{16} \), we convert \( \frac{3}{8} \) to sixteenths:
\[ \frac{3}{8} = \frac{6}{16} \]
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 \( \frac{7}{16} \).

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