Probability
Classical Definition of Probability
Grade 12
Question:
<p>Which of the following cannot be valid assignment of probabilities for outcomes of sample space \(S = \{\omega_1, \omega_2, \omega_3, \omega_4, \omega_5, \omega_6, \omega_7\}\)?</p><table border='1'><tr><th>Assignment</th><th>\(\omega_1\)</th><th>\(\omega_2\)</th><th>\(\omega_3\)</th><th>\(\omega_4\)</th><th>\(\omega_5\)</th><th>\(\omega_6\)</th><th>\(\omega_7\)</th></tr><tr><td>(a)</td><td>0.1</td><td>0.01</td><td>0.05</td><td>0.03</td><td>0.01</td><td>0.2</td><td>0.6</td></tr><tr><td>(b)</td><td>\(\frac{1}{7}\)</td><td>\(\frac{1}{7}\)</td><td>\(\frac{1}{7}\)</td><td>\(\frac{1}{7}\)</td><td>\(\frac{1}{7}\)</td><td>\(\frac{1}{7}\)</td><td>\(\frac{1}{7}\)</td></tr><tr><td>(c)</td><td>0.1</td><td>0.2</td><td>0.3</td><td>0.4</td><td>0.5</td><td>0.6</td><td>0.7</td></tr><tr><td>(d)</td><td>-0.1</td><td>0.2</td><td>0.3</td><td>0.4</td><td>-0.2</td><td>0.1</td><td>0.3</td></tr><tr><td>(e)</td><td>\(\frac{1}{14}\)</td><td>\(\frac{2}{14}\)</td><td>\(\frac{3}{14}\)</td><td>\(\frac{4}{14}\)</td><td>\(\frac{5}{14}\)</td><td>\(\frac{6}{14}\)</td><td>\(\frac{15}{14}\)</td></tr></table>
<p>(a) and (b)</p>
<p>(b) and (d)</p>
<p>(c), (d) and (e)</p>
<p>(a) and (c)</p>
Step-by-Step Solution
Key Concept: A valid probability assignment requires two conditions: (1) each probability must be non-negative (≥0), and (2) the sum of all probabilities must equal exactly 1. Check both conditions for each assignment.
<p><strong>Step 1: Check Assignment (a)</strong></p><p>Sum = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1.0 ✓</p><p>All probabilities ≥ 0 ✓ → <strong>VALID</strong></p><p><strong>Step 2: Check Assignment (b)</strong></p><p>Sum = 1/7 + 1/7 + 1/7 + 1/7 + 1/7 + 1/7 + 1/7 = 7/7 = 1 ✓</p><p>All probabilities ≥ 0 ✓ → <strong>VALID</strong></p><p><strong>Step 3: Check Assignment (c)</strong></p><p>Sum = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 ✗</p><p>Sum ≠ 1, violates axiom → <strong>INVALID</strong></p><p><strong>Step 4: Check Assignment (d)</strong></p><p>Contains -0.1 and -0.2 (negative values) ✗</p><p>Violates non-negativity axiom → <strong>INVALID</strong></p><p><strong>Step 5: Check Assignment (e)</strong></p><p>Sum = (1+2+3+4+5+6+15)/14 = 36/14 ≠ 1 ✗</p><p>Last term is 15/14 > 1, making total > 1 → <strong>INVALID</strong></p><p><strong>∴ Answer: (c), (d), and (e) are NOT valid probability assignments</strong></p>
Correct Answer: (c), (d) and (e)