If \( E \) and \( F \) are two events such that \( P(E)>0 \) and \( P(F) \neq 1 \), then \( P(E \,|\, F) \) is:
Show Hint
To solve conditional probability problems, you may need to use the inclusion-exclusion principle and properties of complements to express the probability in different terms.
The formula for conditional probability is given by:
\[
P(E \,|\, F) = \frac{P(E \cap F)}{P(F)}.
\]
We are asked to find the expression for \( P(E \,|\, F) \) in terms of other probabilities.
Using the inclusion-exclusion principle, we know:
\[
P(E \cup F) = P(E) + P(F) - P(E \cap F)
\]
So, we can rearrange this to express \( P(E \cap F) \) as:
\[
P(E \cap F) = P(E \cup F) - P(F).
\]
Substituting this into the formula for conditional probability:
\[
P(E \,|\, F) = \frac{P(E \cup F) - P(F)}{P(F)}.
\]
Now, consider the complement of \( F \), i.e., \( P(\bar{F}) \), and the formula for the conditional probability of the complement of \( E \), \( P(\bar{E} \,|\, F) \). We obtain the final expression for \( P(E \,|\, F) \) in terms of other events:
\[
P(E \,|\, F) = \frac{1 - P(E \cup F)}{P(\bar{F})}.
\]
Thus, the correct answer is option (4).
