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

The objective is to determine \( P(\overline{E} / F) \), the conditional probability that event \( E \) does not occur, given that event \( F \) has occurred.

1. Apply Conditional Probability Definition:
The definition of conditional probability states:
\( P(\overline{E} / F) = \frac{P(\overline{E} \cap F)}{P(F)} \)

2. Utilize the Complement Rule:
Event \( F \) can be expressed as the union of two mutually exclusive events: \( F = (E \cap F) \cup (\overline{E} \cap F) \).
Therefore, the probability of \( F \) is:
\( P(F) = P(E \cap F) + P(\overline{E} \cap F) \).
Rearranging this yields:
\( P(\overline{E} \cap F) = P(F) - P(E \cap F) \)

3. Substitute into Conditional Formula:
Substituting the expression for \( P(\overline{E} \cap F) \) into the conditional probability formula:
\( P(\overline{E} / F) = \frac{P(F) - P(E \cap F)}{P(F)} = 1 - \frac{P(E \cap F)}{P(F)} \).
Recognizing that \( \frac{P(E \cap F)}{P(F)} = P(E / F) \) by the definition of conditional probability, we get:
\( P(\overline{E} / F) = 1 - P(E / F) \)

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

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