Write the values of `a` for which the following distribution of probabilities becomes a probability distribution:
`X=x_i :\ -2 \ \ -1 \ \ 0 \ \ 1`
`P(X=x_i):(1-a)/4 \ \ (1+2a)/4 \ \ (1-2a)/4 \ \ (1+a)/4 \ \ `

Answer

To determine the values of \( a \) for which the given distribution becomes a valid probability distribution, we need to ensure that the sum of all probabilities equals 1 and that each individual probability is non-negative.

### Step 1: Write down the probabilities
The probabilities given are:
- \( P(X = -2) = \frac{1-a}{4} \)
- \( P(X = -1) = \frac{1+2a}{4} \)
- \( P(X = 0) = \frac{1-2a}{4} \)
- \( P(X = 1) = \frac{1+a}{4} \)

### Step 2: Set up the equation for the sum of probabilities
We need to ensure that the sum of these probabilities equals 1:
\[ \frac{1-a}{4} + \frac{1+2a}{4} + \frac{1-2a}{4} + \frac{1+a}{4} = 1 \]

### Step 3: Combine the probabilities
Combining the fractions, we have:
\[ \frac{(1-a) + (1+2a) + (1-2a) + (1+a)}{4} = 1 \]
This simplifies to:
\[ \frac{4}{4} = 1 \]
Thus, the equation holds true, indicating that the sum of probabilities is indeed 1.

### Step 4: Check for non-negativity of each probability
Next, we need to ensure that each probability is non-negative:
1. \( \frac{1-a}{4} \geq 0 \) implies \( 1-a \geq 0 \) or \( a \leq 1 \)
2. \( \frac{1+2a}{4} \geq 0 \) implies \( 1+2a \geq 0 \) or \( a \geq -\frac{1}{2} \)
3. \( \frac{1-2a}{4} \geq 0 \) implies \( 1-2a \geq 0 \) or \( a \leq \frac{1}{2} \)
4. \( \frac{1+a}{4} \geq 0 \) implies \( 1+a \geq 0 \) or \( a \geq -1 \)

### Step 5: Combine the inequalities
Now, we combine the inequalities:
- From \( a \leq 1 \)
- From \( a \geq -\frac{1}{2} \)
- From \( a \leq \frac{1}{2} \)
- From \( a \geq -1 \)

The most restrictive inequalities are:
\[ -\frac{1}{2} \leq a \leq \frac{1}{2} \]

### Conclusion
Thus, the values of \( a \) for which the given distribution becomes a valid probability distribution are:
\[ a \in \left[-\frac{1}{2}, \frac{1}{2}\right] \]