Let A and B be two events in a sample space such that 0 < P(A) < 1 and 0 < P(B) < 1. Which… — GRE General Test

Let $A$ and $B$ be two events in a sample space such that $0 < P(A) < 1$ and $0 < P(B) < 1$ . Which of the following statements must be true? Select all that apply.

If $A$ and $B$ are mutually exclusive, then $A$ and $B$ cannot be independent.Answer
If $P(A \cup B) = P(A) + P(B) - P(A)P(B)$ , then $A$ and $B$ are independent events.Answer
C
If $A$ and $B$ are independent events, then their complements $A^c$ and $B^c$ must be mutually exclusive.
If $P(A|B) > P(A)$ , then $P(B|A) > P(B)$ .Answer
E
If $A$ and $B$ are independent events, then $P(A \cup B) = P(A) + P(B)$ .

Answer

The statements asserting that mutually exclusive non-impossible events cannot be independent, that

P(A \cup B) = P(A) + P(B) - P(A)P(B)

implies independence, and that

P(A|B) > P(A)

implies

P(B|A) > P(B)

are all true.

For events with probabilities strictly between 0 and 1: (1) Mutual exclusivity requires

P(A \cap B) = 0

, whereas independence requires

P(A \cap B) = P(A)P(B) > 0

, so mutually exclusive events cannot be independent. (2) Substituting

P(A \cup B) = P(A) + P(B) - P(A)P(B)

into the addition rule yields

P(A \cap B) = P(A)P(B)

, which defines independence. (3)

P(A|B) > P(A)

is mathematically equivalent to

P(A \cap B) > P(A)P(B)

, which in turn is equivalent to

P(B|A) > P(B)

Step-by-Step Solution

Analyze the relationship between mutual exclusivity and independence.

Mutually exclusive events satisfy

P(A \cap B) = 0

. For independent events,

P(A \cap B) = P(A)P(B)

. Since

P(A) > 0

and

P(B) > 0

P(A)P(B) > 0 \neq 0

. Thus, mutually exclusive non-impossible events can never be independent.

To evaluate structural compatibility between mutual exclusivity and independence.

Apply the addition rule of probability to check the union equation.

General addition rule:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

. Comparing to the given

P(A \cup B) = P(A) + P(B) - P(A)P(B)

shows

P(A \cap B) = P(A)P(B)

, which is the exact condition for independence.

To verify if the given expression for union probability forces independence.

Evaluate the joint probability of complementary events

A^c

and

B^c

Independence of

A

and

B

implies independence of

A^c

and

B^c

. Thus

P(A^c \cap B^c) = (1-P(A))(1-P(B)) > 0

. Because the joint probability is positive, the complements are not mutually exclusive.

To test whether independence of events implies mutual exclusivity of their complements.

Examine the symmetry of conditional probability inequalities.

P(A|B) > P(A) \implies P(A \cap B) > P(A)P(B) \implies P(B|A) = \frac{P(A \cap B)}{P(A)} > P(B)

To evaluate directional dependence between conditional probabilities.

Check the simple addition rule for independent events.

Simple addition

P(A \cup B) = P(A) + P(B)

applies only when

P(A \cap B) = 0

. Independent events have

P(A \cap B) = P(A)P(B) > 0

, so

P(A \cup B) < P(A) + P(B)

To distinguish between addition rules for mutually exclusive vs. independent events.

Key Concept

Theoretical relationships between independent, dependent, mutually exclusive, and conditional events.

Estimated Time:3m 0s

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