Two events A and B are defined on a sample space such that P(A) = 0.60 and P(B) = 0.75. Wh… — GRE General Test

Two events $A$ and $B$ are defined on a sample space such that $P(A) = 0.60$ and $P(B) = 0.75$ . Which of the following statements must be true? Select all such statements.

Events $A$ and $B$ cannot be mutually exclusive.Answer
The probability that both events $A$ and $B$ occur, $P(A \cap B)$ , is at least $0.35$ .Answer
C
Events $A$ and $B$ must be independent.
The conditional probability $P(A \mid B)$ is at least $\frac{7}{15}$ .Answer
E
The maximum possible value of $P(A \text{ or } B)$ is $0.90$ .

Answer

The statements asserting that events

A

and

B

cannot be mutually exclusive, that the joint probability

P(A \cap B)

is at least

0.35

, and that the conditional probability

P(A \mid B)

is at least

\frac{7}{15}

are all correct.

The sum of the probabilities of events

A

and

B

(

1.35

) exceeds

1

, making mutual exclusivity impossible. The inclusion-exclusion principle dictates

P(A \cap B) \geq 0.60 + 0.75 - 1.00 = 0.35

. Consequently, the minimum conditional probability

P(A \mid B)

\frac{0.35}{0.75} = \frac{7}{15}

Step-by-Step Solution

Evaluate mutual exclusivity

A

and

B

were mutually exclusive,

P(A \cap B) = 0

, so

P(A \cup B) = P(A) + P(B) = 0.60 + 0.75 = 1.35

. Since probability cannot exceed

1

, the events cannot be mutually exclusive.

Verify if the sum of individual probabilities exceeds 1.

Determine the minimum joint probability

P(A \cap B)

Using

P(A \cup B) = P(A) + P(B) - P(A \cap B) \leq 1

, we have

0.60 + 0.75 - P(A \cap B) \leq 1 \implies P(A \cap B) \geq 0.35

Apply the inclusion-exclusion principle bounded by maximum total probability.

Test for required independence

Independence requires

P(A \cap B) = 0.60 \times 0.75 = 0.45

. Since

P(A \cap B)

can legitimately range anywhere between

0.35

and

0.60

, independence is possible but not guaranteed.

Check whether joint probability is strictly fixed at the product of individual probabilities.

Calculate the lower bound for conditional probability

P(A \mid B)

P(A \mid B) = \frac{P(A \cap B)}{P(B)} \geq \frac{0.35}{0.75} = \frac{35}{75} = \frac{7}{15}

Substitute the minimum joint probability into the conditional probability formula.

Key Concept

Probability rules governing overlap, mutual exclusivity, joint probability bounds, and conditional probability.

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic