Probability
Total Probability Theorem
Grade 12
Question:
<p>An event \(X\) can take place in conjuction with any one of the mutually exclusive and exhaustive events \(A\), \(B\) and \(C\). If \(A\), \(B\), \(C\) are equiprobable and the probability of \(X\) is 5/12, and the probability of \(X\) taking place when \(A\) has happened is 3/8, while it is 1/4 when \(B\) has taken place, then the probability of \(X\) taking place in conjuction with \(C\) is</p>
<p>(1) 5/8</p>
<p>(2) 3/8</p>
<p>(3) 5/24</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: Use the law of total probability: P(X) = P(X|A)P(A) + P(X|B)P(B) + P(X|C)P(C). Since A, B, C are equiprobable and exhaustive, P(A) = P(B) = P(C) = 1/3, allowing you to solve for the unknown P(X|C).
<p><strong>Step 1:</strong> Since A, B, C are mutually exclusive and exhaustive with equiprobability, we have P(A) = P(B) = P(C) = 1/3.</p><p><strong>Step 2:</strong> Apply the law of total probability: P(X) = P(X|A)·P(A) + P(X|B)·P(B) + P(X|C)·P(C)</p><p><strong>Step 3:</strong> Substitute known values: 5/12 = (3/8)·(1/3) + (1/4)·(1/3) + P(X|C)·(1/3)</p><p><strong>Step 4:</strong> Simplify: 5/12 = 3/24 + 1/12 + P(X|C)·(1/3) = 1/8 + 1/12 + P(X|C)·(1/3)</p><p><strong>Step 5:</strong> Find common denominator for 1/8 + 1/12: 1/8 + 1/12 = 3/24 + 2/24 = 5/24</p><p><strong>Step 6:</strong> Therefore: 5/12 = 5/24 + P(X|C)·(1/3), which gives P(X|C)·(1/3) = 5/12 - 5/24 = 10/24 - 5/24 = 5/24</p><p><strong>Step 7:</strong> Solve: P(X|C) = (5/24)·3 = 15/24 = 5/8</p><p><strong>Note:</strong> If the question asks for P(X∩C), then P(X∩C) = P(X|C)·P(C) = (5/8)·(1/3) = 5/24. If asking specifically for probability in conjunction (joint probability), verify the exact wording. The answer 1 suggests a reframed question context.</p>
Correct Answer: 1