What is the probability drawing 4 cards so that at least one card doesn't belong to the set S = {2, 6, J, K}, given one card belongs to the set?

Question

Four cards are drawn without replacement from a standard playing deck of 52 cards.

We can calculate $Pr($2 is in the draw $)=1-Pr($2 is not in the draw$)=1-\frac{_{48}C_4}{_{52}C_4}$, but I have no idea how to proceed further.

Answer 1

The total number of hands: $$\binom{52}4$$ The number of hands with no cards belonging to the set $S=\{\text{twos, sixes, jacks, kings}\}$: $$\binom{52-16}4=\binom{36}4$$ The number of hands with at least one card in $S$: $$\binom{52}4-\binom{36}4$$ The number of hands with at least one card in $S$ and at least one card not in $S$: $$\binom{52}4-\binom{36}4-\binom{16}4$$ The conditional probability of "at least one card not in $S$" given "at least one card in $S$": $$\frac{\binom{52}4-\binom{36}4-\binom{16}4}{\binom{52}4-\binom{36}4}=\frac{210000}{211820}=\frac{1500}{1513}$$