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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?

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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?

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The 9 can be drawn

  • before the first 8
  • between first and second 8
  • after second 8

All 3 possibilities have the same probability. Therefore the chance that 9 appears later than both 8s is $\frac{1}{3}$.

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"All 3 possibilities have the same probability"? why? – nbubis Oct 31 '12 at 14:37
Among all 10! orderings of the 10 cards, there are as many orderings in case 1, 2 and 3. – wnvl Oct 31 '12 at 14:54
cfr this question:… – wnvl Oct 31 '12 at 14:55

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