A=2, 4, 6, B=1, 3, 5. What is the probability that the sum of two numbers is at least 8?

The sets are $A = \{2, 4, 6\}$ and $B = \{1, 3, 5\}$.
The number of elements in set A is $n(A) = 3$.
The number of elements in set B is $n(B) = 3$.
The total number of possible pairs when choosing one number from A and one from B is the product of the number of elements in each set.

The possible pairs $(a, b)$ where $a \in A$ and $b \in B$ are:
$(2,1), (2,3), (2,5)$
$(4,1), (4,3), (4,5)$
$(6,1), (6,3), (6,5)$

We are looking for pairs where the sum is "at least 8", which means the sum is greater than or equal to 8.
Let's list the sums:
$2+1=3$
$2+3=5$
$2+5=7$
$4+1=5$
$4+3=7$
$4+5=9$ (This sum is $\ge 8$)
$6+1=7$
$6+3=9$ (This sum is $\ge 8$)
$6+5=11$ (This sum is $\ge 8$)

The pairs whose sum is at least 8 are $(4,5)$, $(6,3)$, and $(6,5)$.
So, the number of favorable outcomes is $n(E) = 3$.

The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes.

The probability that the sum of two numbers is at least 8 is $\boxed{\frac{1}{3}}$.