A 60-card deck contains 11 marked cards. Five cards are drawn as an opening hand and set aside without being looked at. Three more cards are then drawn at random from the remaining 55. What is the probability that at least one of those three cards is marked?

Step-by-step solution

1. 1

   Deal with the opening hand first — it is the whole point of the problem. It is tempting to condition on how many marked cards went into the first five. But nothing was observed, and the deck was shuffled uniformly, so the eight cards drawn in total (five, then three) are just a uniformly random ordered sample of eight. By exchangeability, positions 6, 7 and 8 are as likely to be any particular triple as positions 1, 2 and 3.

2. 2

   Reduce to a plain 3-card draw. The question therefore becomes:

   $P(\text{at least one marked among 3 cards drawn from the full }60)$

   with $11$ marked and $60-11=49$ unmarked cards.

3. 3

   Use the complement. "At least one" is a union of overlapping events, so count the single easy complement instead:

   $P(\text{at least one})=1-P(\text{none marked})=1-\frac{\binom{49}{3}}{\binom{60}{3}}$

4. 4

   Compute the two binomial coefficients.

   $\binom{49}{3}=\frac{49\cdot 48\cdot 47}{6}=18424,\qquad \binom{60}{3}=\frac{60\cdot 59\cdot 58}{6}=34220$

5. 5

   Subtract.

   $P=1-\frac{18424}{34220}=\frac{34220-18424}{34220}=\frac{15796}{34220}$

6. 6

   Convert to a percentage.

   $\frac{15796}{34220}=0.461601\ldots\approx 46.2\%$

   Sanity check on the exchangeability step: the answer does not depend on the size of the opening hand at all — drawing $0$, $5$ or $20$ cards first and then $3$ gives the same $46.16\%$, as long as none of them is revealed.