What is the chance to get two pairs from a hand of five cards if I have extra card (♠️ king, ♠️ queen and ♦️ queen) added to the standard cards What is the chance to get two pairs if I have extra card (♠️ king, ♠️ queen and ♦️ queen) added to the standard cards, thus I have 55 cards? Can anybody help me !!
Is my solution right?
$$\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1} + \binom{13}{2}\binom{5}{2}\binom{5}{2}\binom{11}{1}\binom{5}{1} + \binom{13}{2}\binom{6}{2}\binom{6}{2}\binom{11}{1}\frac{\binom{6}{1}}{\binom{55}{5}}$$
First, I chose $2$ ranks of $13$ and then $2$ suits of $4$ (this is for hearts and clovers). Then I chose $2$ ranks of $13$ and then $2$ suits of $5$ suits (this is for diamonds, but I am not sure). Then I chose $2$ ranks of $13$ and then $2$ suits of $6$ suits (this is for spades, but I am not sure).
Lastly, I divided all on $\binom{55}{5}$

Your approach seems fine but there are a few mistakes.
Number of ways to choose two pairs -
i) A pair of queen and a pair of king
$\displaystyle \binom{6}{2}\binom{5}{2}\binom{44}{1}$
ii) A pair of queen and a pair from other $11$ ranks (not king)
$\displaystyle \binom{6}{2} \binom{11}{1} \binom{4}{2} \binom{45}{1}$
iii) A pair of king and a pair from other $11$ ranks (not queen)
$\displaystyle \binom{5}{2} \binom{11}{1} \binom{4}{2} \binom{46}{1}$
iv) Two pairs from other $11$ ranks (other than queen and king)
$\displaystyle \binom{11}{2} \binom{4}{2} \binom{4}{2} \binom{47}{1}$
Now the probability is sum of $(i), (ii), (iii), (iv)$ divided by $\displaystyle {55 \choose 5}$.
