Understanding the Probability Problem

The question asks us to find the probability of the intersection of events B and C, denoted as \(P(B \cap C)\). We are given the probabilities of event B, and the probabilities of specific intersections involving events A, B, and the complement of C (\(\overline{C}\)).

Given Probabilities

• \(P\left( B \right) = \frac{3}{4}\)
• \(P\left( {A \cap B \cap \overline{C}} \right) = \frac{1}{3}\)
• \(P\;\left( {\overline{A}\; \cap \;B\; \cap \;\overline{C}} \right) = \frac{1}{3}\)

Analyzing the Events and Their Relationships

Let's look at the events \(A \cap B \cap \overline{C}\) and \(\overline{A} \cap B \cap \overline{C}\). Both of these events are subsets of \(B \cap \overline{C}\). The key observation is that these two events are mutually exclusive because they involve the occurrence of \(A\) and the non-occurrence of \(A\) (\(\overline{A}\)) respectively, while the other conditions (B and \(\overline{C}\)) are the same for both.

The union of these two mutually exclusive events covers all possibilities where events B and \(\overline{C}\) occur, partitioned by whether A occurs or not. Therefore, the event \(B \cap \overline{C}\) can be expressed as the union of \(A \cap B \cap \overline{C}\) and \(\overline{A} \cap B \cap \overline{C}\).

Mathematically, this is written as:

\(B \cap \overline{C} = (A \cap B \cap \overline{C}) \cup (\overline{A} \cap B \cap \overline{C})\)

Since they are mutually exclusive, the probability of their union is the sum of their individual probabilities:

\(P(B \cap \overline{C}) = P(A \cap B \cap \overline{C}) + P(\overline{A} \cap B \cap \overline{C})\)

Calculating \(P(B \cap \overline{C})\)

Using the given probabilities, we can calculate \(P(B \cap \overline{C})\):

\(P(B \cap \overline{C}) = \frac{1}{3} + \frac{1}{3}\)

\(P(B \cap \overline{C}) = \frac{2}{3}\)

Relating \(P(B)\), \(P(B \cap C)\), and \(P(B \cap \overline{C})\)

Now, consider event B. Event B can be partitioned into two mutually exclusive events based on whether C occurs or not. These two events are \(B \cap C\) (B and C occur) and \(B \cap \overline{C}\) (B occurs and C does not occur).

So, event B is the union of these two mutually exclusive events:

\(B = (B \cap C) \cup (B \cap \overline{C})\)

Since they are mutually exclusive, the probability of B is the sum of the probabilities of these two events:

\(P(B) = P(B \cap C) + P(B \cap \overline{C})\)

Solving for \(P(B \cap C)\)

We are given \(P(B) = \frac{3}{4}\) and we calculated \(P(B \cap \overline{C}) = \frac{2}{3}\). We can substitute these values into the equation:

\(\frac{3}{4} = P(B \cap C) + \frac{2}{3}\)

To find \(P(B \cap C)\), we rearrange the equation:

\(P(B \cap C) = \frac{3}{4} - \frac{2}{3}\)

To subtract the fractions, we find a common denominator, which is 12:

\(P(B \cap C) = \frac{3 \times 3}{4 \times 3} - \frac{2 \times 4}{3 \times 4}\)

\(P(B \cap C) = \frac{9}{12} - \frac{8}{12}\)

\(P(B \cap C) = \frac{9 - 8}{12}\)

\(P(B \cap C) = \frac{1}{12}\)

Conclusion

Based on the given probabilities and the properties of mutually exclusive events and unions, the probability of \(B \cap C\) is \(\frac{1}{12}\).

Revision Table: Key Probability Concepts

Additional Information: Set Theory and Probability

Probability theory often uses concepts from set theory to describe events and their relationships. Events are treated as sets, and sample space is the universal set. The operations on events (union, intersection, complement) correspond directly to set operations.

• The union of events E and F (\(E \cup F\)) corresponds to the union of sets.
• The intersection of events E and F (\(E \cap F\)) corresponds to the intersection of sets.
• The complement of event E (\(\overline{E}\)) corresponds to the complement of set E relative to the sample space.
• Mutually exclusive events correspond to disjoint sets (their intersection is the empty set).

Understanding these set theory connections helps visualize and solve probability problems, especially those involving multiple events like in this question where we dealt with combinations of A, B, and C.