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: Use the definition of conditional probability.
We know that \[ P(\overline{A}\mid \overline{B}) = \frac{P(\overline{A}\cap \overline{B})}{P(\overline{B})} \]

Step 2: Apply De Morgan’s law.
Using De Morgan’s law, \[ \overline{A}\cap \overline{B} = \overline{A\cup B} \] Therefore, \[ P(\overline{A}\cap \overline{B}) = P(\overline{A\cup B}) \]

Step 3: Use complement rule.
\[ P(\overline{A\cup B}) = 1-P(A\cup B) \] Hence, \[ P(\overline{A}\mid \overline{B}) = \frac{1-P(A\cup B)}{P(\overline{B})} \]

Step 4: Final conclusion.
Therefore, \[ \boxed{ \frac{1-P(A\cup B)}{P(\overline{B})} } \]