Question
Suppose you draw 7 cards from a standard deck of 52 cards. What is the probability that you draw three pairs of cards if
- the three pairs must have distinct denominations?
- the three pairs don't necessarily have distinct denominations? (two cases)
Note: Do not include the possibility of having exactly three cards of any denomination.
My attempt
For #1, my solution is $$probability = \frac{{{\binom{4}{2}}^3\binom{13}{3}(46)}}{\binom{52}{7}}$$ And I'm unsure what #2 could be.
I'd appreciate any help!
For second,
We either choose $3$ denominations or $2$. For pairs from $3$ denominations, it is same as the first and for pairs from $2$ denominations, we first choose the denominations and then we choose which we will have $4$ cards of and which $2$ ($2$ ways do so).
So, probability $\displaystyle = \frac{\binom{13}{3} \ {{\binom{4}{2}}^3 \cdot 40} + \binom{13}{2} \cdot 2 \cdot {\binom{4}{2}} \cdot 44}{\binom{52}{7}}$
Also for first, instead of $46$ choices for the $7$th card, you should have only $40$ choices (as $12$ cards cannot be considered - no three cards can have the same denomination).