Question

If \(A\) and \(B\) are any two events of a random experiment, then evaluate \[ P\Big[(A\cap B^c)\cup(A^c\cap B)\cup(A\cap B)\Big] \]

• A. \(P(A)+P(B)\)
• B. \(P(A^c\cup B^c)\)
• C. \(1-P(A\cup B)\)
• D. \(P(A\cup B)\)

Hint:

Whenever probability expressions contain combinations such as \(A\cap B^c\), \(A^c\cap B\), and \(A\cap B\), try drawing a Venn diagram. In most cases these regions combine to form \(A\cup B\), which simplifies the problem immediately.

Answer 1

The Correct Option is D

Concept: In probability theory, whenever complicated combinations of events involving intersections and unions are given, the first step is to simplify the event expression using the laws of set theory. The most important laws used in such questions are:
• Union of disjoint regions forms a complete event region.
• Complement law: \[ A\cap B^c=\text{elements in A but not in B} \]
• Union law: \[ A\cup B=(A\cap B^c)\cup(A^c\cap B)\cup(A\cap B) \]
• Probability of equivalent events is always equal. The main idea here is to identify what region of the Venn diagram the given expression represents and then rewrite it in simplified form.

Step 1: Observe the given probability expression carefully.
We are given the expression \[ P\Big[(A\cap B^c)\cup(A^c\cap B)\cup(A\cap B)\Big] \] Inside the probability function, there are three different event regions joined together by union. The three regions are: \[ A\cap B^c \] which means outcomes belonging to \(A\) but not belonging to \(B\). Second region: \[ A^c\cap B \] which means outcomes belonging to \(B\) but not belonging to \(A\). Third region: \[ A\cap B \] which means outcomes common to both \(A\) and \(B\). Thus all three possible regions connected with events \(A\) and \(B\) are included.

Step 2: Use set theoretic decomposition of union of two events.
We know an important identity from set algebra: \[ A\cup B=(A\cap B^c)\cup(A^c\cap B)\cup(A\cap B) \] This identity states that if we divide the union of two sets into mutually exclusive parts, then it consists exactly of these three disjoint regions. Comparing with the expression given in the question, we observe that the expression inside probability is exactly equal to: \[ A\cup B \] Hence we can rewrite the probability expression as \[ P\Big[(A\cap B^c)\cup(A^c\cap B)\cup(A\cap B)\Big] = P(A\cup B) \]

Step 3: Determine the final answer from the simplified expression.
Since the given event simplifies exactly into the union of events \(A\) and \(B\), therefore the probability becomes \[ P(A\cup B) \] Looking at the options provided, this corresponds to option (4). Hence the required value is \[ \boxed{P(A\cup B)} \]