Analyzing Probability of Events A and B

The problem provides us with certain probabilities related to two events, A and B, and asks us to determine if these events are equally likely, mutually exclusive, or independent based on the given information.

We are given the following probabilities:

• Probability of the complement of the union of A and B: \(P(\overline{A \cup B}) = \dfrac{1}{6}\)
• Probability of the intersection of A and B: \(P(A \cap B) = \dfrac{1}{4}\)
• Probability of the complement of event A: \(P(\overline{A}) = \dfrac{1}{4}\)

We need to use these values to find \(P(A)\) and \(P(B)\) and then check the conditions for equally likely, mutually exclusive, and independent events.

Calculating Probabilities P(A) and P(B)

First, let's find the probability of event A, \(P(A)\). We know that \(P(A) = 1 - P(\overline{A})\).

\(P(A) = 1 - P(\overline{A}) = 1 - \dfrac{1}{4} = \dfrac{4 - 1}{4} = \dfrac{3}{4}\)

So, \(P(A) = \dfrac{3}{4}\).

Next, let's find the probability of the union of A and B, \(P(A \cup B)\). We know that \(P(\overline{A \cup B}) = 1 - P(A \cup B)\).

\(P(A \cup B) = 1 - P(\overline{A \cup B}) = 1 - \dfrac{1}{6} = \dfrac{6 - 1}{6} = \dfrac{5}{6}\)

So, \(P(A \cup B) = \dfrac{5}{6}\).

Now we can find the probability of event B, \(P(B)\), using the formula for the probability of the union of two events:

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

Substitute the values we know:

\(\dfrac{5}{6} = \dfrac{3}{4} + P(B) - \dfrac{1}{4}\)

Combine the terms on the right side:

\(\dfrac{5}{6} = \dfrac{3}{4} - \dfrac{1}{4} + P(B)\)

\(\dfrac{5}{6} = \dfrac{2}{4} + P(B)\)

\(\dfrac{5}{6} = \dfrac{1}{2} + P(B)\)

Solve for \(P(B)\):

\(P(B) = \dfrac{5}{6} - \dfrac{1}{2}\)

To subtract, find a common denominator, which is 6:

\(P(B) = \dfrac{5}{6} - \dfrac{1 \times 3}{2 \times 3} = \dfrac{5}{6} - \dfrac{3}{6} = \dfrac{5 - 3}{6} = \dfrac{2}{6} = \dfrac{1}{3}\)

So, \(P(B) = \dfrac{1}{3}\).

Checking Properties of Events A and B

Now that we have \(P(A)\) and \(P(B)\), we can check if the events A and B are equally likely, mutually exclusive, or independent.

Equally Likely Events

Events A and B are equally likely if \(P(A) = P(B)\).

We found \(P(A) = \dfrac{3}{4}\) and \(P(B) = \dfrac{1}{3}\).

Since \(\dfrac{3}{4} \neq \dfrac{1}{3}\), events A and B are not equally likely.

Mutually Exclusive Events

Events A and B are mutually exclusive if they cannot happen at the same time, which means their intersection is empty. In terms of probability, this means \(P(A \cap B) = 0\).

We are given \(P(A \cap B) = \dfrac{1}{4}\).

Since \(P(A \cap B) = \dfrac{1}{4} \neq 0\), events A and B are not mutually exclusive.

Independent Events

Events A and B are independent if the occurrence of one does not affect the probability of the other. In terms of probability, this means \(P(A \cap B) = P(A) \times P(B)\).

We have \(P(A \cap B) = \dfrac{1}{4}\).

Let's calculate \(P(A) \times P(B)\):

\(P(A) \times P(B) = \dfrac{3}{4} \times \dfrac{1}{3} = \dfrac{3 \times 1}{4 \times 3} = \dfrac{3}{12} = \dfrac{1}{4}\)

Since \(P(A \cap B) = \dfrac{1}{4}\) and \(P(A) \times P(B) = \dfrac{1}{4}\), we see that \(P(A \cap B) = P(A) \times P(B)\).

Therefore, events A and B are independent.

Conclusion

Based on our calculations:

• Events A and B are not equally likely (\(P(A) \neq P(B)\)).
• Events A and B are not mutually exclusive (\(P(A \cap B) \neq 0\)).
• Events A and B are independent (\(P(A \cap B) = P(A) \times P(B)\)).

Thus, events A and B are independent but not equally likely.

Revision Table: Probability Concepts

Additional Information: Probability Rules

Here are some fundamental probability rules used in solving this problem:

• Complement Rule: The probability of an event not happening is 1 minus the probability of it happening. \(P(\overline{A}) = 1 - P(A)\)
• Union Rule: The probability of either event A or event B (or both) occurring is the sum of their individual probabilities minus the probability of their intersection. \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
• De Morgan's Laws (for probability): The probability of the complement of the union is 1 minus the probability of the union. \(P(\overline{A \cup B}) = 1 - P(A \cup B)\). Also, the probability of the complement of the intersection is 1 minus the probability of the intersection. \(P(\overline{A \cap B}) = 1 - P(A \cap B)\). These are helpful when dealing with complements of combined events.

Understanding these basic rules is crucial for solving probability problems involving multiple events.