Probability
Conditional Probability
Grade 12
Question:
<p>All the jacks, queens, kings, and aces of a regular 52 cards deck are taken out. The 16 cards are thoroughly shuffled and my opponent, a person who always tells the truth, simultaneously draws two cards at random and says, "I hold at least one ace". The probability that he holds two aces is</p>
<p>(1) 2/8</p>
<p>(2) 4/9</p>
<p>(3) 2/3</p>
<p>(4) 1/9</p>
Step-by-Step Solution
Key Concept: Use conditional probability: find P(two aces | at least one ace) = P(both aces AND at least one ace) / P(at least one ace). The event 'both aces' is a subset of 'at least one ace', so the numerator simplifies to P(both aces).
<p><strong>Step 1:</strong> Identify the setup. We have 16 cards: 4 aces and 12 non-aces (4 jacks, 4 queens, 4 kings). Two cards are drawn. We need P(both aces | at least one ace).</p><p><strong>Step 2:</strong> Calculate total ways to draw 2 cards from 16: C(16,2) = 120</p><p><strong>Step 3:</strong> Calculate P(at least one ace). It's easier to use complementary: P(at least one ace) = 1 - P(no aces) = 1 - C(12,2)/C(16,2) = 1 - 66/120 = 54/120</p><p><strong>Step 4:</strong> Calculate P(both aces AND at least one ace) = P(both aces) = C(4,2)/C(16,2) = 6/120</p><p><strong>Step 5:</strong> Apply conditional probability formula: P(both aces | at least one ace) = P(both aces AND at least one ace) / P(at least one ace) = (6/120) / (54/120) = 6/54 = 1/9</p><p>∴ Answer: D</p>
Correct Answer: D