Let A, B and C be three mutually exclusive events such that P(A) = `p_(1)`, P(B) = `p_(2)` and P( C) = `p_(3)`. Then,
Let `p_(1) = 1/2(1-p), p_(2) = 1/3(1+2p) " and " p_(3) = 1/5(2+3p)`, then p belongs to

Options

A. `phi`
B. (-1, 2)
C. `(-1,1/47)`
D. `(-1/3,2/3)`

Answer

To solve the problem, we need to find the range of \( p \) given the probabilities of mutually exclusive events \( A \), \( B \), and \( C \).

### Step 1: Write down the probabilities
We are given:
- \( P(A) = p_1 = \frac{1}{2}(1 - p) \)
- \( P(B) = p_2 = \frac{1}{3}(1 + 2p) \)
- \( P(C) = p_3 = \frac{1}{5}(2 + 3p) \)

### Step 2: Use the property of mutually exclusive events
Since \( A \), \( B \), and \( C \) are mutually exclusive, we have:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) \]
This means:
\[ P(A) + P(B) + P(C) \leq 1 \]

### Step 3: Substitute the probabilities into the inequality
Substituting the expressions for \( p_1 \), \( p_2 \), and \( p_3 \):
\[ \frac{1}{2}(1 - p) + \frac{1}{3}(1 + 2p) + \frac{1}{5}(2 + 3p) \leq 1 \]

### Step 4: Simplify the inequality
First, we find a common denominator for the fractions, which is 30:
\[ \frac{15(1 - p)}{30} + \frac{10(1 + 2p)}{30} + \frac{6(2 + 3p)}{30} \leq 1 \]
Combining the fractions gives:
\[ \frac{15(1 - p) + 10(1 + 2p) + 6(2 + 3p)}{30} \leq 1 \]

### Step 5: Expand and combine like terms
Expanding the numerator:
\[ 15 - 15p + 10 + 20p + 12 + 18p \leq 30 \]
Combining like terms:
\[ (15 + 10 + 12) + (-15p + 20p + 18p) \leq 30 \]
This simplifies to:
\[ 37 + 23p \leq 30 \]

### Step 6: Solve for \( p \)
Subtract 37 from both sides:
\[ 23p \leq 30 - 37 \]
\[ 23p \leq -7 \]
Dividing by 23:
\[ p \leq -\frac{7}{23} \]

### Step 7: Find lower bounds for \( p \)
Now we need to ensure that each probability \( p_1 \), \( p_2 \), and \( p_3 \) is non-negative.

1. For \( p_1 \):
\[ \frac{1}{2}(1 - p) \geq 0 \implies 1 - p \geq 0 \implies p \leq 1 \]

2. For \( p_2 \):
\[ \frac{1}{3}(1 + 2p) \geq 0 \implies 1 + 2p \geq 0 \implies p \geq -\frac{1}{2} \]

3. For \( p_3 \):
\[ \frac{1}{5}(2 + 3p) \geq 0 \implies 2 + 3p \geq 0 \implies p \geq -\frac{2}{3} \]

### Step 8: Combine the inequalities
From the above inequalities, we have:
- \( p \leq -\frac{7}{23} \)
- \( p \geq -\frac{1}{2} \)
- \( p \geq -\frac{2}{3} \)

The most restrictive lower bound is \( p \geq -\frac{1}{2} \).

### Step 9: Final range for \( p \)
Thus, we combine the bounds:
\[ -\frac{1}{2} \leq p \leq -\frac{7}{23} \]

### Conclusion
The final range for \( p \) is:
\[ p \in \left[-\frac{1}{2}, -\frac{7}{23}\right] \]