Q.Three cards are drawn from a pack of 52 cards. Find the chance that a) two are queen cards and one is an ace card b) at least one is a diamond card c) all are from the same suit d) they are a king, a queen and a jack.

$n(S)=\binom{52}{3}=22100$.

a) Two queens, one ace: $\binom{4}{2}\times\binom{4}{1}=6\times4=24$, $P=24/22100=6/5525$.

b) At least one diamond, via the complement (no diamond, all 3 from the other 39 cards): $P(\text{no diamond})=\binom{39}{3}/\binom{52}{3}=9139/22100$, so $P(\text{at least one})=1-9139/22100=12961/22100=997/1700$.

c) All same suit: $4\times\binom{13}{3}=4\times286=1144$, $P=1144/22100=22/425$. …

Combinations for direct-count parts (a, c, d); complement (1 minus 'no diamond') for the 'at least one' part (b), since direct counting of …

Trying to directly add up the cases '1 diamond', '2 diamonds', '3 diamonds' for part (b) instead of using …