There are $10$ cards on which numbers from $1$ to $9$ are written. Number $8$ is written twice, other numbers - only once. Cards are pulled in the random order.
What is the probability that $9$ appears later than both $8$s?
There are $10$ cards on which numbers from $1$ to $9$ are written. Number $8$ is written twice, other numbers - only once. Cards are pulled in the random order.
What is the probability that $9$ appears later than both $8$s?
Hints:
The cards other than 8's and the 9 are irrelevant here. Ignore them.
Think of one 8 as red, the other as black, if that helps clarify the problem.
Now, with these three cards, how many permutations are possible when you draw them? And in what proportion of those permutations does the 9 appear last?
The 9 can be drawn
All 3 possibilities have the same probability. Therefore the chance that 9 appears later than both 8s is $\frac{1}{3}$.