White and green left after the last black marble

An urn holds $200$ black, $400$ white and $600$ green marbles, which are removed one at a time at random without replacement. What is the probability that, at the moment the final black marble is removed, the urn still contains at least one white marble and at least one green marble?

$7/12$. The very last marble must be white ($1/3$) or green ($1/2$). If it is green, we also need the last of the black-or-white marbles to be white ($2/3$); if it is white, the last black-or-green must be green ($3/4$). So $\tfrac12\cdot\tfrac23+\tfrac13\cdot\tfrac34=\tfrac{7}{12}$.