[Combinatorics]
Problem: Six tickets numbered 1 through 6 are placed in a box. Two tickets are randomly selected and removed together. What is the probability that the smaller of the two numbers on the tickets selected is less than or equal to 4 ?

Solution: The possible pairs of numbers on the tickets are (listed as ordered pairs): (1,2), (1,3), $(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6)$, and $(5,6)$.

There are fifteen such pairs. (We treat the pair of tickets numbered 2 and 4 as being the same as the pair numbered 4 and 2.)

The pairs for which the smaller of the two numbers is less than or equal to 4 are $(1,2)$, $(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5)$, and $(4,6)$. There are fourteen such pairs.

Therefore, the probability of selecting such a pair of tickets is $\frac{14}{15}$.
We find the probability that the smaller number on the two tickets is NOT less than or equal to 4.

Therefore, the smaller number on the two tickets is at least 5.

Thus, the pair of numbers must be 5 and 6 , since two distinct numbers less than or equal to 6 are being chosen.

The possible pairs of numbers on the tickets are (listed as ordered pairs): (1,2), (1,3), $(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6)$, and $(5,6)$.

There are fifteen such pairs. (We treat the pair of tickets numbered 2 and 4 as being the same as the pair numbered 4 and 2.)

Therefore, the probability that the smaller number on the two tickets is NOT less than or equal to 4 is $\frac{1}{15}$, so the probability that the smaller number on the two tickets IS less than or equal to 4 is $1-\frac{1}{15}=\frac{14}{15}$.

Final answer: $\frac{14}{15}$