Solving a Probability Problem with Intersecting and Complement Events

This problem involves using fundamental probability concepts like intersection, complement, and union of events to find the value of an unknown variable, \(p\). We are given the probability of the intersection of two events A and B, the probability of the complement of their intersection (which is equivalent to the complement of their union by De Morgan's Law), and a relationship between the probabilities of events A and B, expressed in terms of \(p\).

Understanding the Given Information

We are provided with the following probabilities and relations:

• \(P(A \cap B) = \dfrac{1}{2}\): The probability that both event A and event B occur.
• \(P(\bar A \cap \bar B) = \dfrac{1}{2}\): The probability that neither event A nor event B occurs.
• \(2 P(A) = P(B) = p\): This gives us a relationship between the probabilities of events A and B, and the variable \(p\). From this, we can write:
  • \(P(B) = p\)
  • \(2 P(A) = p \implies P(A) = \dfrac{p}{2}\)

Applying Probability Rules

The term \(P(\bar A \cap \bar B)\) can be simplified using De Morgan's Law, which states that \( \bar A \cap \bar B = \overline{A \cup B} \). This means the event "neither A nor B" is the same as the event "not (A or B)".

So, \(P(\bar A \cap \bar B) = P(\overline{A \cup B})\).

We also know that the probability of the complement of an event E is \(P(\overline{E}) = 1 - P(E)\). Applying this to \(A \cup B\), we get:

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

From the given information, \(P(\bar A \cap \bar B) = \dfrac{1}{2}\). Therefore,

\(1 - P(A \cup B) = \dfrac{1}{2}\)

Solving for \(P(A \cup B)\):

\(P(A \cup B) = 1 - \dfrac{1}{2} = \dfrac{1}{2}\)

Using the Formula for Union of Events

The general formula for the probability of the union of two events A and B is:

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

We have values or expressions for all terms in this equation:

• \(P(A \cup B) = \dfrac{1}{2}\) (calculated above)
• \(P(A) = \dfrac{p}{2}\) (from the given relation)
• \(P(B) = p\) (from the given relation)
• \(P(A \cap B) = \dfrac{1}{2}\) (given)

Substitute these values into the union formula:

\(\dfrac{1}{2} = \dfrac{p}{2} + p - \dfrac{1}{2}\)

Solving for p

Now, we need to solve the equation for \(p\):

Combine the constant terms:

\(\dfrac{1}{2} + \dfrac{1}{2} = \dfrac{p}{2} + p\)

\(1 = \dfrac{p}{2} + p\)

Combine the terms involving \(p\). To add \(\dfrac{p}{2}\) and \(p\), write \(p\) as \(\dfrac{2p}{2}\):

\(1 = \dfrac{p}{2} + \dfrac{2p}{2}\)

\(1 = \dfrac{p + 2p}{2}\)

\(1 = \dfrac{3p}{2}\)

Multiply both sides by 2:

\(1 \times 2 = 3p\)

\(2 = 3p\)

Divide both sides by 3:

\(p = \dfrac{2}{3}\)

Conclusion

The value of \(p\) that satisfies the given conditions is \(\dfrac{2}{3}\).

Let's check if this value makes sense. If \(p = \dfrac{2}{3}\), then \(P(B) = \dfrac{2}{3}\) and \(P(A) = \dfrac{p}{2} = \dfrac{2/3}{2} = \dfrac{2}{6} = \dfrac{1}{3}\). Since probabilities must be between 0 and 1 (inclusive), these values are valid.

The calculated value of p is \(\dfrac{2}{3}\), which matches option 3.

Revision Table: Key Probability Formulas Used

Additional Information: Exploring Related Concepts

Understanding probability involves several core concepts. Here are a few related to the problem we just solved:

• Sample Space: The set of all possible outcomes of a random experiment. Probabilities are assigned to subsets of the sample space (events).
• Event: A subset of the sample space.
• Mutually Exclusive Events: Two events A and B are mutually exclusive (or disjoint) if they cannot occur at the same time. Their intersection is the empty set, so \(P(A \cap B) = 0\). In this case, the union formula simplifies to \(P(A \cup B) = P(A) + P(B)\).
• Independent Events: Two events A and B are independent if the occurrence of one does not affect the probability of the other. Mathematically, this is expressed as \(P(A \cap B) = P(A) \times P(B)\). Independence and mutual exclusivity are distinct concepts; independent events (unless one has probability 0) cannot be mutually exclusive.
• Venn Diagrams: Visual representations of sets and their relationships. They are very useful for understanding union, intersection, and complement of events in probability problems. A Venn diagram could visually show the regions corresponding to \(A \cap B\), \(A \cup B\), \(\bar A \cap \bar B\), etc.

This problem required us to connect the probability of the intersection of complements (\(P(\bar A \cap \bar B)\)) to the probability of the union (\(P(A \cup B)\)) using De Morgan's law and the complement rule, and then use the standard union formula along with the given relationship between \(P(A)\) and \(P(B)\) to solve for the unknown variable \(p\).