A 60-card deck contains 9 marked cards. Three cards are drawn at random without replacement. What is the probability that at least one of them is marked?

Step-by-step solution

1. 1

   Use the complement, not a case split. "At least one marked" covers one, two, or three marked cards. Its complement is the single event "no marked card at all", so

   $P(\text{at least one})=1-P(\text{none})$

   This is always the cheaper route for an at least one question.

2. 2

   Count the total number of three-card draws. Order does not matter, so

   $\binom{60}{3}=\frac{60\times59\times58}{3\times2\times1}=34220$

3. 3

   Count the draws with no marked card. All three must come from the $51$ unmarked cards:

   $\binom{51}{3}=\frac{51\times50\times49}{3\times2\times1}=20825$

4. 4

   Form the complement probability.

   $P(\text{none})=\frac{20825}{34220}=0.60856\ldots$

   so

   $P(\text{at least one})=1-\frac{20825}{34220}=\frac{34220-20825}{34220}=\frac{13395}{34220}$

   Reduced, this is $\tfrac{2679}{6844}$.

5. 5

   Convert to a percentage.

   $\frac{13395}{34220}=0.391438\ldots\approx39.14\%$

6. 6

   Sanity-check against a simpler estimate. Each single draw hits a marked card with probability $\tfrac{9}{60}=15\%$. Three near-independent draws would miss all three with probability roughly $0.85^3=0.614$, giving about $38.6\%$ — very close to the exact $39.14\%$, with the small gap coming from drawing without replacement ✓.

7. 7

   Note why the opening hand does not matter. If five cards were already dealt face-down and ignored, the three cards drawn afterwards are still a uniformly random three-card subset of the original $60$, so the same computation applies unchanged.