Question

For two mutually exclusive events

• A. \(P(A \cup B) = P(A) + P(B)\)
• B. \(P(A \cup B) = P(A) \times P(B)\)
• C. \(P(A \cup B) = \frac{P(A) + P(B)}{P(A \cap B)}\)
• D. \(P(A \cup B) = P(A) - P(B)\)

Hint:

For mutually exclusive events, think of non-overlapping Venn diagrams.
Since they do not overlap, there is no shared region to subtract, so we simply add the probabilities.

Answer 1

The Correct Option is A

Step 1: Recall what mutually exclusive means.
Two events are mutually exclusive when they cannot both happen on the same trial, which means their intersection is the empty set, so \( P(A \cap B) = 0 \).
Step 2: Apply the general addition rule.
For any two events, the general rule is \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). This rule always holds, regardless of whether the events are exclusive or not.
Step 3: Simplify for the exclusive case.
Since \( P(A \cap B) = 0 \) here, the general rule reduces directly to \( P(A \cup B) = P(A) + P(B) \), with no subtraction term left over.
\[ \boxed{P(A \cup B) = P(A) + P(B)} \]