Question

If \( E \) and \( F \) are two events such that \( P(E)>0 \) and \( P(F) \neq 1 \), then \( P(E \,|\, F) \) is:

• A. \( \frac{P(\bar{E})}{P(\bar{F})} \)
• B. \( 1 - P(\bar{E} \,|\, F) \)
• C. \( \frac{1 - P(E \cup F)}{P(\bar{F})} \)
• D. \( 1 - P(E / F) \)

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.

Answer 1

The Correct Option is D

To solve the problem, we need to find the value of \( P(\overline{E} / F) \), i.e., the conditional probability of the complement of event \( E \), given event \( F \) has occurred.

1. Use the Definition of Conditional Probability:
By definition:
\( P(\overline{E} / F) = \frac{P(\overline{E} \cap F)}{P(F)} \)

2. Use the Complement Rule:
We know that:
\( F = (E \cap F) \cup (\overline{E} \cap F) \), and these two events are mutually exclusive.
So:
\( P(F) = P(E \cap F) + P(\overline{E} \cap F) \)
Therefore:
\( P(\overline{E} \cap F) = P(F) - P(E \cap F) \)

3. Substitute Back into the Conditional Formula:
\( P(\overline{E} / F) = \frac{P(F) - P(E \cap F)}{P(F)} = 1 - \frac{P(E \cap F)}{P(F)} \)
Using the definition of conditional probability again:
\( \frac{P(E \cap F)}{P(F)} = P(E / F) \)
So:
\( P(\overline{E} / F) = 1 - P(E / F) \)

4. Conclusion:
The required value is:
\( P(\overline{E} / F) = 1 - P(E / F) \)

Final Answer:
The correct option is (D) 1 − P(E / F).