Question

If \( E_1 \) and \( E_2 \) are mutually exclusive events, then prove that \( P(E_1) + P(E_2) = P(E_1 \cup E_2) + P(E_1 \cap E_2) \).

Hint:

The general addition rule for any two events is \( P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \). Rearranging this gives \( P(E_1) + P(E_2) = P(E_1 \cup E_2) + P(E_1 \cap E_2) \). This identity is actually true for *any* pair of events, not just mutually exclusive ones. The condition of being mutually exclusive simplifies \( P(E_1 \cap E_2) \) to 0.

Answer 1

Step 1: Understanding the Concept:
We need to prove a relationship between the probabilities of two events, given that they are mutually exclusive. Mutually exclusive events are events that cannot happen at the same time.
Step 2: Key Formula or Approach:
1. Use the definition of mutually exclusive events: If \( E_1 \) and \( E_2 \) are mutually exclusive, their intersection is the empty set, i.e., \( E_1 \cap E_2 = \emptyset \). 2. Use the axiom of probability for the empty set: \( P(\emptyset) = 0 \). 3. Use the addition axiom for mutually exclusive events: \( P(E_1 \cup E_2) = P(E_1) + P(E_2) \). 4. Manipulate these facts to arrive at the desired equation.
Step 3: Detailed Explanation:
The proof is based on the axioms of probability and the definition of mutually exclusive events.
By definition, if events \( E_1 \) and \( E_2 \) are mutually exclusive, then they cannot occur simultaneously. This means their intersection is the null or empty event. \[ E_1 \cap E_2 = \emptyset \] The probability of the empty event is zero. \[ P(E_1 \cap E_2) = P(\emptyset) = 0 \] One of the fundamental axioms of probability (the addition rule for mutually exclusive events) states that: \[ P(E_1 \cup E_2) = P(E_1) + P(E_2) \] This equation holds true precisely because the events are mutually exclusive.
Now, let's look at the equation we need to prove: \[ P(E_1) + P(E_2) = P(E_1 \cup E_2) + P(E_1 \cap E_2) \] We can start from the addition axiom for mutually exclusive events: \[ P(E_1) + P(E_2) = P(E_1 \cup E_2) \] We can add 0 to the right side of the equation without changing its value: \[ P(E_1) + P(E_2) = P(E_1 \cup E_2) + 0 \] Since we have already established that \( P(E_1 \cap E_2) = 0 \) for mutually exclusive events, we can substitute this into the equation: \[ P(E_1) + P(E_2) = P(E_1 \cup E_2) + P(E_1 \cap E_2) \] This completes the proof.
Step 4: Final Answer:
Using the definition and axioms related to mutually exclusive events, we have proven that the given identity holds true.