Probability
Classical Probability
Grade 12

Question:

<p>A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. Find the probability that the determinant chosen is non-zero.</p>
<p>\(\dfrac{3}{16}\)</p>
<p>\(\dfrac{3}{8}\)</p>
<p>\(\dfrac{1}{4}\)</p>
<p>\(\dfrac{1}{8}\)</p>

Step-by-Step Solution

Key Concept: Count total 2×2 matrices with entries {0,1}, then count those with non-zero determinants using the condition |A| = ad - bc ≠ 0.
<p><strong>Step 1:</strong> Count total possible 2×2 matrices with entries from {0,1}.</p><p>A 2×2 matrix has 4 entries, each can be 0 or 1.</p><p>Total determinants = 2⁴ = 16</p><p><strong>Step 2:</strong> For matrix <code>⎡a b⎤</code>, det = ad - bc</p><p><code>⎣c d⎦</code></p><p>Det is non-zero when ad ≠ bc</p><p><strong>Step 3:</strong> Count cases where ad - bc = 0 (det = 0):</p><p>• If a=0 or d=0: ad=0, need bc=0 → b=0 or c=0 → 2⁴ - 12 = 4 cases where ad=bc=0</p><p>• If ad=1 (a=d=1): need bc=1 → b=c=1 → 1 case</p><p>• If ad=0 and bc=0: cases are (a,d,b,c) where ad=bc. Systematic count: 12 cases give det=0</p><p><strong>Step 4:</strong> Direct count of non-zero determinants:</p><p>• a=1, d=1, b=0, c=0: det = 1 ✓</p><p>• a=1, d=1, b=0, c=1: det = 1 ✓</p><p>• a=1, d=1, b=1, c=0: det = 1 ✓</p><p>• a=0, d=1, b=0, c=1: det = -0 ✗</p><p>• a=1, d=0, b=1, c=0: det = -1 ✓</p><p>Counting systematically: 4 matrices give non-zero determinant</p><p><strong>Step 5:</strong> Probability = (Non-zero determinants)/(Total determinants) = 4/16 = 1/4</p><p>∴ Answer: <strong>1/4</strong></p>
Correct Answer: B

Master Probability with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free