Suppose `E_1, E_2 and E_3` be three mutually exclusive events such that `P(E_i)=p_i" for "i=1, 2, 3`.
`P(E_1capoverline(E_2))+P(E_2capoverline(E_3))+P(E_3capoverline(E_1))` equals

Options

A. (a)`p_1(1-p_2)+p_2(1-P_3)+p_3(1-p_1)`
B. (b)`p_1p_2+p_2p_3+p_3p_1`
C. (c)0
D. (d)None of these

Answer

To solve the problem, we need to find the value of the expression:

\[ P(E_1 \cap \overline{E_2}) + P(E_2 \cap \overline{E_3}) + P(E_3 \cap \overline{E_1}) \]

where \( E_1, E_2, \) and \( E_3 \) are mutually exclusive events with probabilities \( P(E_1) = p_1, P(E_2) = p_2, P(E_3) = p_3 \).

### Step-by-step Solution:

1. **Understanding the Events**:
Since \( E_1, E_2, \) and \( E_3 \) are mutually exclusive, it means that the occurrence of one event excludes the occurrence of the others. Therefore, if \( E_1 \) occurs, neither \( E_2 \) nor \( E_3 \) can occur.

2. **Calculating \( P(E_1 \cap \overline{E_2}) \)**:
The event \( E_1 \cap \overline{E_2} \) means \( E_1 \) occurs and \( E_2 \) does not occur. Since \( E_2 \) cannot occur if \( E_1 \) occurs (due to mutual exclusivity), we have:
\[ P(E_1 \cap \overline{E_2}) = P(E_1) = p_1 \]

3. **Calculating \( P(E_2 \cap \overline{E_3}) \)**:
Similarly, for \( E_2 \cap \overline{E_3} \):
\[ P(E_2 \cap \overline{E_3}) = P(E_2) = p_2 \]

4. **Calculating \( P(E_3 \cap \overline{E_1}) \)**:
For \( E_3 \cap \overline{E_1} \):
\[ P(E_3 \cap \overline{E_1}) = P(E_3) = p_3 \]

5. **Combining the Results**:
Now we can combine these results:
\[ P(E_1 \cap \overline{E_2}) + P(E_2 \cap \overline{E_3}) + P(E_3 \cap \overline{E_1}) = p_1 + p_2 + p_3 \]

6. **Considering the Total Probability**:
Since \( E_1, E_2, \) and \( E_3 \) are mutually exclusive and collectively exhaustive events, we have:
\[ p_1 + p_2 + p_3 = 1 \]

7. **Final Result**:
Therefore, we conclude that:
\[ P(E_1 \cap \overline{E_2}) + P(E_2 \cap \overline{E_3}) + P(E_3 \cap \overline{E_1}) = 1 \]

### Conclusion:
The final answer is \( 1 \).