Question 3

Mathematics Probability Medium

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

(A) \( \frac{1}{4}, \frac{1}{2} \)
(B) \( \frac{1}{2}, \frac{1}{4} \)
(C) \( \frac{1}{2}, 1 \)
(D) \( 1, \frac{1}{2} \)
View Dynamic Solution & Explanation
Correct Solution: Option D

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: \( {1, \frac{1}{2}} \)