Question

If $ A $ and $ B $ are two events such that $ P(B) \ne 0 $ and $ P(B) \ne 1 $, then: $$ P(\bar{A} \mid B) = ? $$

• A. \( 1 - P(A \mid B) \)
• B. \( 1 - P(\bar{A} \mid B) \)
• C. \( \frac{1 - P(A \cup B)}{P(B)} \)
• D. \( \frac{P(\bar{A})}{P(B)} \)

Hint:

N/A

Answer 1

The Correct Option is C

We know:

\[ P(\overline{A} \mid B) = \frac{P(\overline{A} \cap B)}{P(B)} \]

Now:

\[ P(\overline{A} \cap B) = P(B) - P(A \cap B) \Rightarrow P(\overline{A} \mid B) = \frac{P(B) - P(A \cap B)}{P(B)} = 1 - \frac{P(A \cap B)}{P(B)} = 1 - P(A \mid B) \]

This gives:

\[ \boxed{1 - P(A \mid B)} \]

Wait! But the marked answer is option 3: \(\frac{1 - P(A \cup B)}{P(B)}\)

Let's analyze it:

\[ P(\overline{A} \mid B) = \frac{P(B \cap \overline{A})}{P(B)} = \frac{P(B) - P(A \cap B)}{P(B)} = 1 - P(A \mid B) \]

So correct answer is:

\[ \boxed{1 - P(A \mid B)} \Rightarrow \text{Option (1) is actually correct} \]

But per the image answer, marked is Option 3 — which is:

\[ \frac{1 - P(A \cup B)}{P(B)} \Rightarrow \text{This is not a valid expression for } P(\overline{A} \mid B) \]

Accurate result: Option 1 is correct.