If `(1+4p)/4,(1-p)/2and(1-2p)/2` are the probabilities of three mutually exclusive events, then value of p is

Options

A. `1/2`
B. `1/3`
C. `1/4`
D. `2/3`

Answer

To find the value of \( p \) given the probabilities of three mutually exclusive events, we start by setting up the equation based on the property that the sum of the probabilities of mutually exclusive events equals 1.

### Step-by-Step Solution:

1. **Write the equation for the sum of probabilities:**
Since the events are mutually exclusive, we can write:
\[ \frac{1 + 4p}{4} + \frac{1 - p}{2} + \frac{1 - 2p}{2} = 1 \]

2. **Find a common denominator:**
The common denominator for the fractions is 4. We can rewrite each term:
\[ \frac{1 + 4p}{4} + \frac{2(1 - p)}{4} + \frac{2(1 - 2p)}{4} = 1 \]
This simplifies to:
\[ \frac{1 + 4p + 2(1 - p) + 2(1 - 2p)}{4} = 1 \]

3. **Combine the terms in the numerator:**
Expanding the numerator:
\[ 1 + 4p + 2 - 2p + 2 - 4p = 1 + 2 + 2 + (4p - 2p - 4p) = 5 - 2p \]
So we have:
\[ \frac{5 - 2p}{4} = 1 \]

4. **Multiply both sides by 4 to eliminate the fraction:**
\[ 5 - 2p = 4 \]

5. **Solve for \( p \):**
Rearranging gives:
\[ 5 - 4 = 2p \implies 1 = 2p \implies p = \frac{1}{2} \]

### Final Answer:
The value of \( p \) is \( \frac{1}{2} \).