Adjudicating.
Label the cards \(\displaystyle R_{1},R_{2},R_{3},B_{1},B_{2},B_{3}.\)
Give person number 1 two random cards, 6C2 ways of doing that.
Give person number 2 two of the remaining four cards, 4C2 ways of doing that.
Person number 3 gets what's left.
The total number of ways of doing that is 6C2*4C2 = 15*6 = 90.
Now person number 1 is to have 1 red card and 1 blue card. There are 3*3 = 9 ways in which that can happen.
Person number 2 is to have 1 red card and 1 blue card from the remaining 4 cards. That can happen in 2*2 = 4 ways.
Person number 3 has what's left.
So, the number of ways in which each person has 1 red card and 1 blue card is 9*4 = 36.
So, the probability of each person has 1 red card and 1 blue card is 36/90 = 2/5.