Let \( P(E) \) denote the probability of event \( E \) . Given \( P(A)=1, P(B) \) \( =1 / 2 \) , the values of \( P(A \mid B) \) and \( P(B \mid A) \) respectively are
Step-by-step Solution:
We are given:
\( P(A) = 1 \)
\( P(B) = \frac{1}{2} \)
We are to find:
\( P(A \mid B) \) and
\( P(B \mid A) \)
Step 1: Use conditional probability formulas
Formula 1:
\[
P(A \mid B) = \frac{P(A \cap B)}{P(B)}
\]
Formula 2:
\[
P(B \mid A) = \frac{P(A \cap B)}{P(A)}
\]
Step 2: Use the given data
Since \( P(A) = 1 \), it means event \( A \) always occurs.
So the intersection \( P(A \cap B) = P(B) \), because if B happens, A must already be happening.
So:
\( P(A \cap B) = P(B) = \frac{1}{2} \)
Now:
\( P(A \mid B) \):
\[
P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1
\]
\( P(B \mid A) \):
\[
P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{2}}{1} = \frac{1}{2}
\]
So the correct option is:
Option D: \( {1, \frac{1}{2}} \)