Probability
Classical Probability
Grade 12
Question:
<p>A card is drawn and replaced in an ordinary pack of 52 cards. The minimum number of times a card must be drawn so that the probability of getting at least one ace exceeds \(\dfrac{1}{2}\) is</p>
<p>8</p>
<p>9</p>
<p>7</p>
<p>10</p>
Step-by-Step Solution
Key Concept: P(at least one ace in n draws) > 1/2 ⟺ 1-(48/52)ⁿ > 1/2 ⟺ (12/13)ⁿ < 1/2.
<p>$P(\text{no ace}) = \dfrac{48}{52} = \dfrac{12}{13}$.</p><p>Need: $\left(\dfrac{12}{13}\right)^n < \dfrac{1}{2}$.</p><p>Taking logarithm: $n > \dfrac{\ln 2}{\ln(13/12)} \approx \dfrac{0.6931}{0.0800} \approx 8.66$.</p><p>Minimum $n = 9$.</p><p>Verify: $(12/13)^8 \approx 0.538 > 0.5$ ✗; $(12/13)^9 \approx 0.497 < 0.5$ ✓.</p>
Correct Answer: B