Verified Answer :

  • Mutually exclusive events by definition cannot occur at the same time.
  • This property implies their intersection is the empty set: \(A \cap B = \emptyset\).
  • The probability of an impossible event is \(0\), so \(P(A \cap B) = 0\).
  • The correct option is \(\boxed{b}\).

Explanation

  1. Define mutually exclusive events

In probability theory, two events are considered mutually exclusive if they cannot occur simultaneously. This means that if event \(A\) occurs, event \(B\) cannot occur, and vice versa. In terms of set theory, this is expressed as the intersection of events \(A\) and \(B\) being an empty set, denoted as \(A \cap B = \emptyset\).

  1. Determine the probability of the intersection

Since the intersection \(A \cap B\) represents the event that both \(A\) and \(B\) occur at the same time, and mutually exclusive events cannot both occur, the event \(A \cap B\) is an impossible event. In probability, the probability of an impossible event is zero. Therefore, \(P(A \cap B) = 0\).

  1. Evaluate the options and conclude

Comparing this to the given options:

A\(P(A) + P(B) = 0\) is not necessarily true unless both probabilities are zero.
B\(P(A \cap B) = 0\) matches our derived condition for mutual exclusivity.
C\(P(A \cap B) = 1\) is false, as this would imply the events always occur together.
D\(P(A) + P(B) = 1\) is only true for complementary events (where \(B\) is the complement of \(A\)), which is a specific case of mutual exclusivity but not the general definition.

Thus, option \(b\) is the correct statement.

Examples

Mutually exclusive events are essential in everyday decision-making, such as predicting the outcome of a single-game match where one team must either win, lose, or tie. Since a team cannot win and tie the same game, the events "win" and "tie" are mutually exclusive, and their intersection probability is \(\$0\). Understanding this property allows for accurate summation of probabilities, as the probability of either event occurring is simply the sum of their individual probabilities: \(P(A \cup B) = P(A) + P(B)\), provided that \(P(A \cap B) = 0\).