Probability
Dice probability
Grade 12
Question:
<p><strong>260.</strong> The probability of occurrence of a multiple of 2 on one dice and a multiple of 3 on the other dice if both are thrown together, is:</p>
<p>(a) \(\dfrac{7}{36}\)</p>
<p>(b) \(\dfrac{1}{3}\)</p>
<p>(c) \(\dfrac{1}{6}\)</p>
<p>(d) \(\dfrac{11}{36}\)</p>
Step-by-Step Solution
Key Concept: Use the multiplication rule for independent events: find the probability of getting a multiple of 2 on one die AND a multiple of 3 on the other die, accounting for both possible orderings (M2 on first die with M3 on second, or vice versa).
<p><strong>Step 1:</strong> Identify multiples on a die (1-6).</p><p>Multiples of 2: {2, 4, 6} → P(multiple of 2) = 3/6 = 1/2</p><p>Multiples of 3: {3, 6} → P(multiple of 3) = 2/6 = 1/3</p><p><strong>Step 2:</strong> Since we need a multiple of 2 on ONE die and multiple of 3 on THE OTHER die, we have two cases:</p><p>• Case 1: M2 on first die AND M3 on second die = (1/2) × (1/3) = 1/6</p><p>• Case 2: M3 on first die AND M2 on second die = (1/3) × (1/2) = 1/6</p><p><strong>Step 3:</strong> Add both cases (mutually exclusive events):</p><p>P(total) = 1/6 + 1/6 = 2/6 = <strong>1/3</strong></p><p>∴ Answer: D (1/3)</p>
Correct Answer: D