Question

For two events $A$ and $B$ such that $P(A)>0$ and $P(B)=1$, $P(A^c/B^c)$ is

• A. $1-P(A/B)$
• B. $1-P(A^c/B)$
• C. $\dfrac{1-P(A\cap B)}{P(B)}$
• D. $\dfrac{1-P(A\cup B)}{P(B^c)}$

Hint:

For conditional probability, the complement rule always holds: \(P(A^c/B)=1-P(A/B)\).

Answer 1

The Correct Option is A

Step 1: Recall definition of conditional probability.
Conditional probability is defined as
\[ P(A/B)=\frac{P(A\cap B)}{P(B)} \] where \(P(B)\neq0\).
Step 2: Use complement rule in conditional probability.
For any event \(A\), the complement rule states
\[ P(A^c/B)=1-P(A/B) \] because either event \(A\) occurs or its complement occurs under condition \(B\).
Step 3: Apply the property.
Thus directly we obtain
\[ P(A^c/B)=1-P(A/B) \] Step 4: Interpret the result.
This means that the probability of the complement event given \(B\) equals one minus the probability of the event itself under the same condition.
Hence the correct expression becomes
\[ 1-P(A/B) \] Final Answer: $\boxed{1-P(A/B)}$