Answer:
[tex]\frac{17}{33}[/tex]
Explanation:
The probability of finding at least on pair is equal to 1 minus the probability of finding no pair. If x represents the number of pairs obtained in a game we have that:
[tex]P(x\geq 1)=1-P(x=0)[/tex]
Now lets consider the a game in which no pairs are obtained. The first card can be any, so we have 12 out 12 cards. The second card can't be the one matching the first card, so we have 10 out of 11. Similarly for the 3rd and 4th cards, we obtain:
[tex]P(x=0)=\frac{12}{12} \frac{10}{11} \frac{8}{10} \frac{6}{9} =\frac{16}{33}[/tex]
This results in
[tex]P(x\geq 1)=1-\frac{16}{33}=\frac{17}{33}[/tex]
