Certainly! Let's break down the problem step-by-step.
Given:
- The probability of event [tex]\( A \)[/tex] is [tex]\( x \)[/tex].
- The probability of event [tex]\( B \)[/tex] is [tex]\( y \)[/tex].
- The two events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are independent.
We need to find out which condition must be true:
A. [tex]\( P(A \mid B) = x \)[/tex]
B. [tex]\( P(A \mid B) = y \)[/tex]
C. [tex]\( P(B \mid A) = x \)[/tex]
D. [tex]\( P(B \mid A) = xy \)[/tex]
### Understanding Independence
When two events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are independent, the occurrence of one event does not affect the occurrence of the other. Mathematically, for independent events, the following holds true:
[tex]\[ P(A \cap B) = P(A) \cdot P(B) \][/tex]
### Conditional Probability
The conditional probability of [tex]\( A \)[/tex] given [tex]\( B \)[/tex] is defined as:
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \][/tex]
Since [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are independent, we substitute [tex]\( P(A \cap B) \)[/tex] using the property of independence:
[tex]\[ P(A \cap B) = P(A) \cdot P(B) = x \cdot y \][/tex]
So,
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{x \cdot y}{y} \][/tex]
Provided [tex]\( P(B) = y \neq 0 \)[/tex], this simplifies to:
[tex]\[ P(A \mid B) = x \][/tex]
Thus, for independent events, the conditional probability [tex]\( P(A \mid B) \)[/tex] equals the probability of event [tex]\( A \)[/tex], which is [tex]\( x \)[/tex].
Considering the options:
- Option A: [tex]\( P(A \mid B) = x \)[/tex] — This is true.
- Option B: [tex]\( P(A \mid B) = y \)[/tex] — This is not true.
- Option C: [tex]\( P(B \mid A) = x \)[/tex] — This is not true.
- Option D: [tex]\( P(B \mid A) = xy \)[/tex] — This is not true.
Therefore, the correct answer is:
A. [tex]\( P(A \mid B) = x \)[/tex]
