Probability
Mutually Exclusive Events
Grade 12
Question:
<p><strong>State whether the statements are true or false.</strong><br>The probabilities of the events \(A\), \(B\) and \(C\) are respectively \(\frac{3}{2}\), \(\frac{1}{4}\) and \(\frac{1}{6}\). Then the events are mutually exclusive.</p>
<p>(a) True</p>
<p>(b) False</p>
Step-by-Step Solution
Key Concept: Probability of any event must satisfy 0 ≤ P(E) ≤ 1. Since P(A) = 3/2 > 1, event A is impossible and the statement is invalid before even checking mutual exclusivity.
<p><strong>Step 1:</strong> Recall that for any event E, the probability must satisfy: 0 ≤ P(E) ≤ 1</p><p><strong>Step 2:</strong> Check if P(A) = 3/2 is valid. Since 3/2 = 1.5 > 1, this violates the fundamental axiom of probability.</p><p><strong>Step 3:</strong> Since P(A) itself is not a valid probability, the event A cannot exist in any valid probability space.</p><p><strong>Step 4:</strong> Therefore, the statement claiming these are events with the given probabilities is <strong>FALSE</strong>. The problem setup itself is invalid.</p><p>∴ Answer: <strong>FALSE</strong> (The statement is false because P(A) = 3/2 exceeds 1, which is impossible for any probability.)</p>
Correct Answer: A