Both blue and green appear in four draws with replacement

A box holds six marbles, each a different colour; one is blue and one is green. You draw a marble four times, returning it to the box after each draw. What is the probability that blue and green each appear at least once among the four draws?

$\tfrac{151}{648}$. Use inclusion–exclusion on the complements: no blue has probability $(\tfrac56)^4$, as does no green, and neither has $(\tfrac46)^4$. So the answer is $1-2\cdot\tfrac{625}{1296}+\tfrac{256}{1296}=\tfrac{302}{1296}=\tfrac{151}{648}$.