Question

Let \(A,B,C\) be three events such that \(P(C)P(C^c)>0\). Consider the following statements.

• A. \(S1\) is correct and \(S2\) is NOT correct
• B. \(S1\) is NOT correct and \(S2\) is correct
• C. Both \(S1\) and \(S2\) are correct
• D. Neither \(S1\) nor \(S2\) is correct

Hint:

Use the law of total probability whenever conditional probabilities with respect to \(C\) and \(C^c\) are given together.

Answer 1

The Correct Option is C

Step 1: Set up $S1$ with total probability.
Since $P(C),P(C^c)>0$, write $P(A)=P(A|C)P(C)+P(A|C^c)P(C^c)$ and the same for $B$.

Step 2: Compare term by term.
Given $P(A|C)>P(B|C)$ and $P(A|C^c)>P(B|C^c)$, multiplying by the positive weights and adding gives $P(A)>P(B)$. So $S1$ is correct.

Step 3: Handle $S2$.
Now $P(A|C)=P(A|C^c)=k$. Then $P(A)=kP(C)+kP(C^c)=k$.

Step 4: Read off independence.
So $P(A|C)=P(A)$, which means $P(A\cap C)=P(A)P(C)$, i.e. $A$ and $C$ are independent. $S2$ is correct.

Step 5: Conclude.
Both correct, option (C).
\[ \boxed{(C)} \]