Understanding Independent Events and Probabilities

When dealing with probability, understanding the properties of independent events is crucial. Two events, say A and B, are considered independent if the occurrence of one does not affect the probability of the other occurring. In this problem, we are given that A and B are two independent events, and their probabilities \({\rm{P}}\left( {\rm{A}} \right) \ne 0\) and \({\rm{P}}\left( {\rm{B}} \right) \ne 0\). We are also introduced to their complements, \({\rm{\bar A}}\) and \({\rm{\bar B}}\). Our goal is to identify which of the given statements about these probabilities is False.

Analyzing Statement 1: Probability of Intersection

The first statement is: \({\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right){\rm{\;}} = {\rm{\;P}}\left( {\rm{A}} \right){\rm{P}}\left( {\rm{B}} \right)\).

• Concept: By definition, if two events A and B are independent events, the probability of both events occurring (their intersection) is the product of their individual probabilities. This is a fundamental property of independent events.
• Conclusion: Since A and B are given as independent events, this statement is True.

Analyzing Statement 2: Conditional Probability

The second statement is: \({\rm{P}}\left( {{\rm{A}}/{\rm{B}}} \right){\rm{\;}} = {\rm{\;P}}\left( {\rm{A}} \right)\).

• Concept: The general formula for conditional probability \({\rm{P}}\left( {{\rm{A}}/{\rm{B}}} \right)\) (probability of A given B) is \({\rm{P}}\left( {{\rm{A}}/{\rm{B}}} \right) = \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}}\).
• Since A and B are independent events, we know from Statement 1 that \({\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right) = {\rm{P}}\left( {\rm{A}} \right){\rm{P}}\left( {\rm{B}} \right)\).
• Substituting this into the conditional probability formula: $${\rm{P}}\left( {{\rm{A}}/{\rm{B}}} \right) = \frac{{{\rm{P}}\left( {\rm{A}} \right){\rm{P}}\left( {\rm{B}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}}$$
• Given that \({\rm{P}}\left( {\rm{B}} \right) \ne 0\), we can cancel \({\rm{P}}\left( {\rm{B}} \right)\) from the numerator and denominator: $${\rm{P}}\left( {{\rm{A}}/{\rm{B}}} \right) = {\rm{P}}\left( {\rm{A}} \right)$$
• Conclusion: This statement is True for independent events.

Analyzing Statement 3: Probability of Union

The third statement is: \({\rm{P}}\left( {{\rm{A}} \cup {\rm{B}}} \right){\rm{\;}} = {\rm{\;P}}\left( {\rm{A}} \right){\rm{\;}} + {\rm{\;P}}\left( {\rm{B}} \right)\).

• Concept: The general addition rule for the probability of the union of any two events A and B is: $${\rm{P}}\left( {{\rm{A}} \cup {\rm{B}}} \right) = {\rm{P}}\left( {\rm{A}} \right) + {\rm{P}}\left( {\rm{B}} \right) - {\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)$$
• The given statement \({\rm{P}}\left( {{\rm{A}} \cup {\rm{B}}} \right){\rm{\;}} = {\rm{\;P}}\left( {\rm{A}} \right){\rm{\;}} + {\rm{\;P}}\left( {\rm{B}} \right)\) is only true if events A and B are mutually exclusive (also known as disjoint events), meaning they cannot occur at the same time. In that case, \({\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right) = 0\).
• However, for independent events, we know that \({\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right) = {\rm{P}}\left( {\rm{A}} \right){\rm{P}}\left( {\rm{B}} \right)\).
• Substituting this into the general addition rule for independent events: $${\rm{P}}\left( {{\rm{A}} \cup {\rm{B}}} \right) = {\rm{P}}\left( {\rm{A}} \right) + {\rm{P}}\left( {\rm{B}} \right) - {\rm{P}}\left( {\rm{A}} \right){\rm{P}}\left( {\rm{B}} \right)$$
• Since we are given \({\rm{P}}\left( {\rm{A}} \right) \ne 0\) and \({\rm{P}}\left( {\rm{B}} \right) \ne 0\), it implies that \({\rm{P}}\left( {\rm{A}} \right){\rm{P}}\left( {\rm{B}} \right) \ne 0\).
• Therefore, for independent events with non-zero probabilities, \({\rm{P}}\left( {{\rm{A}} \cup {\rm{B}}} \right)\) is generally less than \({\rm{P}}\left( {\rm{A}} \right) + {\rm{P}}\left( {\rm{B}} \right)\) by the amount \({\rm{P}}\left( {\rm{A}} \right){\rm{P}}\left( {\rm{B}} \right)\).
• Independent events with non-zero probabilities cannot be mutually exclusive. If they were mutually exclusive, \({\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right) = 0\). But if they were also independent, \({\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right) = {\rm{P}}\left( {\rm{A}} \right){\rm{P}}\left( {\rm{B}} \right)\). This would mean \({\rm{P}}\left( {\rm{A}} \right){\rm{P}}\left( {\rm{B}} \right) = 0\), which contradicts the given condition that \({\rm{P}}\left( {\rm{A}} \right) \ne 0\) and \({\rm{P}}\left( {\rm{B}} \right) \ne 0\).
• Conclusion: This statement is False for independent events with non-zero probabilities.

Analyzing Statement 4: Probability of Complements

The fourth statement is: \({\rm{P}}\left( {{\rm{\bar A}} \cap {\rm{\bar B}}} \right){\rm{\;}} = {\rm{\;P}}\left( {{\rm{\bar A}}} \right){\rm{P}}\left( {{\rm{\bar B}}} \right)\).

• Concept: A property of independent events is that if A and B are independent, then their complements \(\bar{A}\) and \(\bar{B}\) are also independent.
• If \(\bar{A}\) and \(\bar{B}\) are independent events, then by the definition of independence, the probability of their intersection is the product of their individual probabilities.
• Proof for clarity: We know \({\rm{P}}\left( {{\rm{\bar A}} \cap {\rm{\bar B}}} \right) = {\rm{P}}\left( {\overline {{\rm{A}} \cup {\rm{B}}} } \right)\) (by De Morgan's Law). Also, \({\rm{P}}\left( {\overline {{\rm{A}} \cup {\rm{B}}} } \right) = 1 - {\rm{P}}\left( {{\rm{A}} \cup {\rm{B}}} \right)\). For independent events A and B, we have \({\rm{P}}\left( {{\rm{A}} \cup {\rm{B}}} \right) = {\rm{P}}\left( {\rm{A}} \right) + {\rm{P}}\left( {\rm{B}} \right) - {\rm{P}}\left( {\rm{A}} \right){\rm{P}}\left( {\rm{B}} \right)\). So, \({\rm{P}}\left( {{\rm{\bar A}} \cap {\rm{\bar B}}} \right) = 1 - \left( {{\rm{P}}\left( {\rm{A}} \right) + {\rm{P}}\left( {\rm{B}} \right) - {\rm{P}}\left( {\rm{A}} \right){\rm{P}}\left( {\rm{B}} \right)} \right)\) \({\rm{P}}\left( {{\rm{\bar A}} \cap {\rm{\bar B}}} \right) = 1 - {\rm{P}}\left( {\rm{A}} \right) - {\rm{P}}\left( {\rm{B}} \right) + {\rm{P}}\left( {\rm{A}} \right){\rm{P}}\left( {\rm{B}} \right)\). Now consider the right side: \({\rm{P}}\left( {{\rm{\bar A}}} \right){\rm{P}}\left( {{\rm{\bar B}}} \right)\). We know \({\rm{P}}\left( {{\rm{\bar A}}} \right) = 1 - {\rm{P}}\left( {\rm{A}} \right)\) and \({\rm{P}}\left( {{\rm{\bar B}}} \right) = 1 - {\rm{P}}\left( {\rm{B}} \right)\). So, \({\rm{P}}\left( {{\rm{\bar A}}} \right){\rm{P}}\left( {{\rm{\bar B}}} \right) = \left( {1 - {\rm{P}}\left( {\rm{A}} \right)} \right)\left( {1 - {\rm{P}}\left( {\rm{B}} \right)} \right)\) \({\rm{P}}\left( {{\rm{\bar A}}} \right){\rm{P}}\left( {{\rm{\bar B}}} \right) = 1 - {\rm{P}}\left( {\rm{B}} \right) - {\rm{P}}\left( {\rm{A}} \right) + {\rm{P}}\left( {\rm{A}} \right){\rm{P}}\left( {\rm{B}} \right)\). Since both sides simplify to the same expression, the statement is true.
• Conclusion: This statement is True.

Identifying the False Statement

Based on our analysis, the only statement that is False for independent events A and B with non-zero probabilities is: $${\rm{P}}\left( {{\rm{A}} \cup {\rm{B}}} \right){\rm{\;}} = {\rm{\;P}}\left( {\rm{A}} \right){\rm{\;}} + {\rm{\;P}}\left( {\rm{B}} \right)$$ This formula incorrectly implies that the independent events are also mutually exclusive, which is not possible when their individual probabilities are not zero.