Understanding Probability with Simultaneous and Exactly One Event

This problem involves calculating probabilities related to the complements and intersections of two events, A and B, based on the given probabilities of their simultaneous occurrence and the occurrence of exactly one of them.

Given Information

• The probability of simultaneous occurrence of two events A and B is p. This means $P(A \cap B) = p$.
• The probability that exactly one of A, B occurs is q. This means $P(\text{exactly one of A, B}) = q$.

Relationship between Events

The probability that exactly one of A, B occurs can be expressed as the probability of A occurring but not B, plus the probability of B occurring but not A. Alternatively, it is the probability of their union minus the probability of their intersection:

$P(\text{exactly one of A, B}) = P(A \cup B) - P(A \cap B) $

Using the given values:

$q = P(A \cup B) - p$

From this, we can find the probability of the union of A and B:

$P(A \cup B) = p + q$

Evaluating Statement 1: $P(A̅) + P(B̅)$

Statement 1 claims that $P(A̅) + P(B̅) = 2 – 2p – q$. Let's evaluate the left side using probability rules.

The probability of the complement of an event E is $P(E̅) = 1 - P(E)$. Applying this to A and B:

• $P(A̅) = 1 - P(A)$
• $P(B̅) = 1 - P(B)$

So, $P(A̅) + P(B̅) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.

We know the formula for the union of two events is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Rearranging this, we get $P(A) + P(B) = P(A \cup B) + P(A \cap B)$.

Substitute the values we derived:

$P(A) + P(B) = (p + q) + p = 2p + q$

Now substitute this back into the expression for $P(A̅) + P(B̅)$:

$P(A̅) + P(B̅) = 2 - (2p + q) = 2 - 2p - q$

This matches Statement 1. Therefore, Statement 1 is correct.

Evaluating Statement 2: $P(A̅ ∩ B̅)$

Statement 2 claims that $P(A̅ ∩ B̅) = 1 – p – q$. Let's evaluate the left side.

Using De Morgan's Law, the intersection of the complements of two events is equivalent to the complement of their union:

$P(A̅ \cap B̅) = P((A \cup B)̅)$

The probability of the complement of an event is 1 minus the probability of the event:

$P((A \cup B)̅) = 1 - P(A \cup B)$

We previously found that $P(A \cup B) = p + q$. Substitute this value:

$P(A̅ \cap B̅) = 1 - (p + q) = 1 - p - q$

This matches Statement 2. Therefore, Statement 2 is also correct.

Conclusion

Both Statement 1 and Statement 2 have been shown to be correct based on the given information and fundamental probability rules.

Revision Table: Probability Concepts

Additional Information: Probability of Events

Probability is a measure of the likelihood that an event will occur. For any event E, the probability $P(E)$ is between 0 and 1, inclusive ($0 \le P(E) \le 1$). A probability of 0 means the event is impossible, and a probability of 1 means the event is certain.

When dealing with multiple events, understanding how they relate to each other is crucial. Key relationships include:

• Intersection ($A \cap B$): Occurs when both A and B happen.
• Union ($A \cup B$): Occurs when A happens, or B happens, or both happen.
• Complement ($A̅$): Occurs when event A does not happen.
• Exactly One: Occurs when one specific event happens, but the other does not. This can be written as $(A \cap B̅) \cup (A̅ \cap B)$.

De Morgan's laws provide a way to relate unions and intersections of complements:

• $(A \cup B)̅ = A̅ \cap B̅$
• $(A \cap B)̅ = A̅ \cup B̅$

These formulas and concepts are fundamental tools for solving various probability problems, including those involving simultaneous events and calculating probabilities of complements and unions.