Question

If \(A\) and \(B\) are two events such that \[ P(B)\neq0 \quad \text{and} \quad P(\overline{B})\neq1, \] then \[ P(\overline{A}\mid \overline{B}) = \]

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

Hint:

Remember: \[ P(X\mid Y)=\frac{P(X\cap Y)}{P(Y)} \] and De Morgan’s law: \[ \overline{A}\cap\overline{B}=\overline{A\cup B}. \] These identities are frequently used together in probability problems.

Answer 1

The Correct Option is C

Step 1: State what we need to find.
We need to express $P(\bar{A}|\bar{B})$ in terms of other probabilities.
Step 2: Apply the definition of conditional probability.
$P(\bar{A}|\bar{B}) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})}$.
Step 3: Express $P(\bar{A} \cap \bar{B})$ using De Morgan's law.
$\bar{A} \cap \bar{B} = \overline{A \cup B}$, so $P(\bar{A} \cap \bar{B}) = 1 - P(A \cup B)$.
Step 4: Substitute back.
$P(\bar{A}|\bar{B}) = \frac{1 - P(A \cup B)}{P(\bar{B})}$.
Step 5: Match with options.
This matches option (3): $\frac{1-P(A \cup B)}{P(\bar{B})}$.
Step 6: Verify the other options are incorrect.
Option (1) $1-P(A|B)$ would require $P(\bar{A}|\bar{B}) + P(A|B) = 1$, which is not generally true. Option (2) $1-P(\bar{A}|B)$ is also not the same expression. Option (4) $\frac{P(\bar{A})}{P(\bar{B})}$ omits the dependency structure. Only option (3) follows directly from the definition.
\[ \boxed{\dfrac{1-P(A \cup B)}{P(\bar{B})}} \]