To find the probability \( P(K = 1 \mid X = \frac{1}{2}) \), we can use Bayes' theorem:
\[ P(K = 1 \mid X = \frac{1}{2}) = \frac{P(X = \frac{1}{2} \mid K = 1) P(K = 1)}{P(X = \frac{1}{2})} \]
Step 1: Determine \( P(K = 1) \)
Since \( K \) can take values 1, 2, and 3 with equal probability, we have:
\[ P(K = 1) = \frac{1}{3} \]
Step 2: Compute \( P(X = \frac{1}{2} \mid K = k) \)
- For \( K = 1 \):
\[ f_{X|K}(x|1) = 1 \quad \text{for } x \in [0, 1] \]
Therefore,
\[ P\left(X = \frac{1}{2} \mid K = 1\right) = 1 \]
- For \( K = 2 \):
\[ f_{X|K}(x|2) = 2x \quad \text{for } x \in [0, 1] \]
Thus,
\[ P\left(X = \frac{1}{2} \mid K = 2\right) = 2 \left( \frac{1}{2} \right) = 1 \]
- For \( K = 3 \):
\[ f_{X|K}(x|3) = 3x^2 \quad \text{for } x \in [0, 1] \]
Therefore,
\[ P\left(X = \frac{1}{2} \mid K = 3\right) = 3 \left( \frac{1}{2} \right)^2 = 3 \cdot \frac{1}{4} = \frac{3}{4} \]
Step 3: Calculate \( P(X = \frac{1}{2}) \)
Using the law of total probability, we can calculate \( P(X = \frac{1}{2}) \):
\[ P(X = \frac{1}{2}) = P(X = \frac{1}{2} \mid K = 1) P(K = 1) + P(X = \frac{1}{2} \mid K = 2) P(K = 2) + P(X = \frac{1}{2} \mid K = 3) P(K = 3) \]
Substituting in the values we computed:
\[ P(X = \frac{1}{2}) = 1 \cdot \frac{1}{3} + 1 \cdot \frac{1}{3} + \frac{3}{4} \cdot \frac{1}{3} \]
Calculating each term:
\[ P(X = \frac{1}{2}) = \frac{1}{3} + \frac{1}{3} + \frac{1}{4} = \frac{2}{3} + \frac{1}{4} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12} \]
Step 4: Substitute everything back into Bayes' theorem
Now we have:
\[ P(K = 1 \mid X = \frac{1}{2}) = \frac{1 \cdot \frac{1}{3}}{\frac{11}{12}} = \frac{\frac{1}{3}}{\frac{11}{12}} = \frac{1}{3} \cdot \frac{12}{11} = \frac{4}{11} \]
Thus, the probability that \( K = 1 \) given that \( X = \frac{1}{2} \) is:
\[ \boxed{\frac{4}{11}} \]