Question
Let K be a discrete random variable that can take the values 1, 2, and 3, all with equal probability Suppose that X takes values in [0, 1] and that for z in that interval we have
f X|K (x|k)= 1,& if k = 1 ,\\ 2 x,&if k=2,\\ 3 x^ 2 ,&if k=3
Find the probability that K = 1 given that X = 1/2
Answers
GPT-4o mini
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 \( P(X = \frac{1}{2} \mid K = k) \)
1. For \( K = 1 \):
\[
f_{X|K}(x|1) = 1 \quad \text{for } x \in [0, 1]
\]
Therefore,
\[
P\left(X = \frac{1}{2} \mid K = 1\right) = 1
\]
2. For \( K = 2 \):
\[
f_{X|K}(x|2) = 2x \quad \text{for } x \in [0, 1]
\]
Thus,
\[
P\left(X = \frac{1}{2} \mid K = 2\right) = 2 \left( \frac{1}{2} \right) = 1
\]
3. 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:
\[
\boxed{\frac{4}{11}}
\]
\[
P(K = 1 \mid X = \frac{1}{2}) = \frac{P(X = \frac{1}{2} \mid K = 1) P(K = 1)}{P(X = \frac{1}{2})}
\]
### 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 \( P(X = \frac{1}{2} \mid K = k) \)
1. For \( K = 1 \):
\[
f_{X|K}(x|1) = 1 \quad \text{for } x \in [0, 1]
\]
Therefore,
\[
P\left(X = \frac{1}{2} \mid K = 1\right) = 1
\]
2. For \( K = 2 \):
\[
f_{X|K}(x|2) = 2x \quad \text{for } x \in [0, 1]
\]
Thus,
\[
P\left(X = \frac{1}{2} \mid K = 2\right) = 2 \left( \frac{1}{2} \right) = 1
\]
3. 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:
\[
\boxed{\frac{4}{11}}
\]