Statement-1 : If `(1)/(5)(1+5p),(1)/(3)(1+2p),(1)/(3)(1-p) " and " (1)/(5)(1-3p)` are probabilities of four mutually exclusive events, then p can take infinite number of values.
Statement-2 : If A, B, C and D are four mutually exclusive events, then
`P(A), P(B), P(C ), P(D) ge 0`
and `P(A)+P(B)+P(C )+P(D) le 1`

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Answer

To solve the problem, we need to analyze the two statements provided and verify their correctness step by step.

### Step 1: Analyze Statement 2

Statement 2 states that if A, B, C, and D are four mutually exclusive events, then:
- \( P(A), P(B), P(C), P(D) \geq 0 \)
- \( P(A) + P(B) + P(C) + P(D) \leq 1 \)

**Verification:**
- Since probabilities cannot be negative, \( P(A), P(B), P(C), P(D) \geq 0 \) is indeed true.
- The sum of probabilities of mutually exclusive events must be less than or equal to 1. Thus, \( P(A) + P(B) + P(C) + P(D) \leq 1 \) is also true.

**Conclusion for Statement 2:** Statement 2 is true.

### Step 2: Analyze Statement 1

Statement 1 provides four expressions representing probabilities of mutually exclusive events:
1. \( \frac{1}{5}(1 + 5p) \)
2. \( \frac{1}{3}(1 + 2p) \)
3. \( \frac{1}{3}(1 - p) \)
4. \( \frac{1}{5}(1 - 3p) \)

We need to ensure that the sum of these probabilities lies between 0 and 1.

**Setting Up the Inequality:**
\[ 0 \leq \frac{1}{5}(1 + 5p) + \frac{1}{3}(1 + 2p) + \frac{1}{3}(1 - p) + \frac{1}{5}(1 - 3p) \leq 1 \]

### Step 3: Calculate the Sum of Probabilities

1. **Combine the terms:**
\[ \frac{1}{5}(1 + 5p) + \frac{1}{3}(1 + 2p) + \frac{1}{3}(1 - p) + \frac{1}{5}(1 - 3p) \]

2. **Distributing the fractions:**
\[ = \frac{1}{5} + p + \frac{1}{3} + \frac{2p}{3} + \frac{1}{3} - \frac{p}{3} + \frac{1}{5} - \frac{3p}{5} \]

3. **Combine like terms:**
- Constant terms: \( \frac{1}{5} + \frac{1}{3} + \frac{1}{3} + \frac{1}{5} \)
- Coefficient of \( p \): \( 1 + \frac{2}{3} - \frac{1}{3} - \frac{3}{5} \)

4. **Finding a common denominator (15):**
- Constant terms:
\[ \frac{3}{15} + \frac{5}{15} + \frac{5}{15} + \frac{3}{15} = \frac{16}{15} \]
- Coefficient of \( p \):
\[ \frac{15}{15} + \frac{10}{15} - \frac{5}{15} - \frac{9}{15} = \frac{11}{15}p \]

5. **Final expression:**
\[ \frac{16}{15} + \frac{11}{15}p \]

### Step 4: Set Up the Inequalities

Now we need to set up the inequalities:
1. \( \frac{16}{15} + \frac{11}{15}p \geq 0 \)
2. \( \frac{16}{15} + \frac{11}{15}p \leq 1 \)

**Solving the First Inequality:**
\[ \frac{11}{15}p \geq -\frac{16}{15} \implies p \geq -\frac{16}{11} \]

**Solving the Second Inequality:**
\[ \frac{11}{15}p \leq 1 - \frac{16}{15} \implies \frac{11}{15}p \leq -\frac{1}{15} \implies p \leq -\frac{1}{11} \]

### Step 5: Conclusion for Statement 1

From the inequalities, we have:
\[ -\frac{16}{11} \leq p \leq -\frac{1}{11} \]
This interval contains infinitely many values for \( p \).

**Conclusion for Statement 1:** Statement 1 is true.

### Final Conclusion

Both statements are true, and Statement 2 correctly explains Statement 1.