Exercise :
Assume you pick randomly $6$ cards from a normal deck of $52$ cards.
(a) What is the probability of all cards having the same color ? (for example, 6 hearts)
(b) What is the probability of picking $3+3$ cards ? (for example 3 dames and 3 10s)
Attempt :
(a)
$$P(a) = \frac{\text{# of ways to have 6 cards that are all hearts }}{\text{# of ways to have 6 cards in general}} = \frac{?}{ \binom{52}{6}} = \dots$$
Is the above elaboration correct ? If so, what is the binomial number for the numerator that I need to set ? I know that there are $13$ hearts in a deck of $52$ cards.
(b)
$$P(b) = \frac{\binom{13}{3}\binom{39}{0}+\binom{13}{3}\binom{39}{0}}{\binom{52}{6}}$$
since we want $3$ out of the $13$ possible types and the rest can be whatever. Is this correct though ?
Sorry if this is elementary but this is my first course of combinatorics/probabilities and having to deal with cards (which I was never interested in) makes it a bit harder.
You can choose all 6 cards from either Hearts or Diamonds or Spades or Clubs each of 13 cards. Therefore there will be $4\binom{13}{6}$ different such ways to do that and the corresponding probability is $$P(a)=\dfrac{4\binom{13}{6}}{\binom{52}{6}}$$for the latter probability we first need to choose two different and distinct 4-tuples (such as dames and 10s) in $\binom{13}{2}$ ways and choose $3$ cards from each set which is possible in $\binom{13}{2}\binom{4}{3}\binom{4}{3}$ different ways and the probability is $$P(b)=\dfrac{\binom{13}{2}\binom{4}{3}\binom{4}{3}}{\binom{52}{6}}$$