Probability
Classical Probability
Grade 12
Question:
<p>Four candidates <em>A</em>, <em>B</em>, <em>C</em> and <em>D</em> have applied for the post in government office. If <em>A</em> is twice as likely to be selected as <em>B</em>, and <em>B</em> and <em>C</em> are given about the same chances of being selected, while <em>C</em> is twice as likely to be selected as <em>D</em>, what are the probabilities that</p><p>(i) <em>C</em> will be selected?</p><p>(ii) <em>A</em> will not be selected?</p>
Step-by-Step Solution
Key Concept: Set up probability ratios from given likelihood relationships, then normalize using the constraint that all probabilities sum to 1. If A:B = 2:1, B:C = 1:1, and C:D = 2:1, express all in terms of a common variable and solve.
<p><strong>Step 1:</strong> Translate the given conditions into probability ratios.</p><p>• A is twice as likely as B: P(A) = 2P(B)</p><p>• B and C are equally likely: P(B) = P(C)</p><p>• C is twice as likely as D: P(C) = 2P(D)</p><p><strong>Step 2:</strong> Express all probabilities in terms of P(D).</p><p>Let P(D) = k</p><p>Then P(C) = 2k (since C is twice as likely as D)</p><p>Then P(B) = 2k (since B = C)</p><p>Then P(A) = 2P(B) = 4k</p><p><strong>Step 3:</strong> Use the normalization condition P(A) + P(B) + P(C) + P(D) = 1.</p><p>4k + 2k + 2k + k = 1</p><p>9k = 1</p><p>k = 1/9</p><p><strong>Step 4:</strong> Calculate the required probabilities.</p><p>• P(C) = 2k = 2/9 ✓</p><p>• P(A) = 4k = 4/9, so P(A will not be selected) = 1 − 4/9 = 5/9 ✓</p><p><strong>Answer:</strong> (i) 2/9, (ii) 5/9</p>
Correct Answer: (i) 2/9, (ii) 5/9