Answer: B
Solution: Solution: Let $\left( a _ { 1 } , a _ { 2 } \right) , \left( b _ { 1 } , b _ { 2 } \right) , \left( c _ { 1 } , c _ { 2 } \right)$ denote three couples respectively.
There are $\left( a _ { 1 } , a _ { 2 } \right) , \left( a _ { 1 } , b _ { 1 } \right) , \left( a _ { 1 } , b _ { 2 } \right) , \left( a _ { 1 } , c _ { 1 } \right) , \left( a _ { 1 } , c _ { 2 } \right)$, $\left( a _ { 2 } , b _ { 1 } \right) , \left( a _ { 2 } , b _ { 2 } \right) , \left( a _ { 2 } , c _ { 1 } \right) , \left( a _ { 2 } , c _ { 2 } \right) , \left( b _ { 1 } , b _ { 2 } \right) , \left( b _ { 1 } , c _ { 1 } \right) , \left( b _ { 1 } , c _ { 2 } \right) , \left( b _ { 2 } , c _ { 1 } \right) , \left( b _ { 2 } , c _ { 2 } \right) , \left( c _ { 1 } , c _ { 2 } \right)$, and $\left( a _ { 1 } , a _ { 2 } \right) , \left( b _ { 1 } , b _ { 2 } \right) , \left( c _ { 1 } , c _ { 2 } \right)$, a total of 15 cases.
$\left( a _ { 2 } , b _ { 1 } \right) , \left( a _ { 2 } , b _ { 2 } \right) , \left( a _ { 2 } , c _ { 1 } \right) , \left( a _ { 2 } , c _ { 2 } \right) , \left( b _ { 1 } , b _ { 2 } \right) , \left( b _ { 1 } , c _ { 1 } \right) , \left( b _ { 1 } , c _ { 2 } \right) , \left( b _ { 2 } , c _ { 1 } \right) , \left( b _ { 2 } , c _ { 2 } \right) , \left( c _ { 1 } , c _ { 2 } \right)$, 15 in total;
Among them, the probability of drawing exactly one couple is $\left( a _ { 1 } , a _ { 2 } \right) , \left( b _ { 1 } , b _ { 2 } \right) , \left( c _ { 1 } , c _ { 2 } \right)$ , 3 in total, and the probability of drawing exactly one couple is $\left( a _ { 1 } , a _ { 2 } \right) , \left( b _ { 1 } , b _ { 2 } \right) , \left( c _ { 1 } , c _ { 2 } \right)$ , 3 in total.
Therefore, the probability of drawing exactly one couple is $P = \frac { 3 } { 15 } = \frac { 1 } { 5 }$.