A 60-card deck contains 5 cards marked A and 23 different cards marked B. Five cards are drawn at random without replacement. What is the probability that the five cards include at least one A and at least one B?

Step-by-step solution

1. 1

   Count the three disjoint groups of cards. The marks do not overlap, so the deck splits into

   $5\ \text{marked A},\qquad 23\ \text{marked B},\qquad 60-5-23=32\ \text{unmarked}$

   Every later count is a choice of $5$ cards from one of these sub-piles, and the total number of equally likely hands is

   $\binom{60}{5}=5{,}461{,}512$

2. 2

   Switch to complements — “at least one … and at least one …” is painful head-on. Counting hands with $1$ A, $2$ A's, $3$ A's … while simultaneously tracking B's means a double sum. Instead let

   $A=\{\text{at least one A}\},\qquad B=\{\text{at least one B}\}$

   and apply inclusion–exclusion to the complement:

   $P(A\cap B)=1-P(A^c\cup B^c)=1-P(A^c)-P(B^c)+P(A^c\cap B^c)$

   The three complementary events each describe hands drawn from a single restricted pile, so each is one binomial coefficient. The $+P(A^c\cap B^c)$ term is essential: hands with neither mark were subtracted twice and must be added back once.

3. 3

   Count hands with no A. These come from the $60-5=55$ non-A cards:

   $P(A^c)=\frac{\binom{55}{5}}{\binom{60}{5}}=\frac{3{,}478{,}761}{5{,}461{,}512}=0.63696$

4. 4

   Count hands with no B. These come from the $60-23=37$ non-B cards:

   $P(B^c)=\frac{\binom{37}{5}}{\binom{60}{5}}=\frac{435{,}897}{5{,}461{,}512}=0.07981$

   That this probability is small is the expected sanity check: with $23$ of $60$ cards marked B, missing all of them in five draws should be unlikely.

5. 5

   Count hands with neither mark. These are drawn from the $32$ unmarked cards only:

   $P(A^c\cap B^c)=\frac{\binom{32}{5}}{\binom{60}{5}}=\frac{201{,}376}{5{,}461{,}512}=0.03687$

6. 6

   Combine. Putting the four counts over the common denominator $\binom{60}{5}$:

   $P(A\cap B)=\frac{5{,}461{,}512-3{,}478{,}761-435{,}897+201{,}376}{5{,}461{,}512}=\frac{1{,}748{,}230}{5{,}461{,}512}$

   $P(A\cap B)=0.32010\approx32.0\%$

7. 7

   Check the result against a rough expectation. The chance of catching at least one A alone is $1-0.63696=0.36304$, and the chance of at least one B alone is $1-0.07981=0.92019$. If the two events were independent the product would be $0.36304\times0.92019=0.33406$; the true answer $0.32010$ is slightly lower, which is exactly right — the events are mildly negatively correlated, because slots in the five-card hand used by an A are slots that cannot hold a B.