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Question:

I have a stack of 30 cards numbered from 1 to 30. I start to pull out of the stack of cards one by one. what is the probability that each one of those series:

$$ 1,2,3,...,10$$ $$11,12,13,...,20$$ $$21,22,23,...,30$$

will not show up in the pulling order.

Answer:

first the probability space size is $\left |\Omega \right |=30!$

Now I want to calculate the probability that one of those series will show up as the complement to the probability I'm looking for.

Let's call the type of series that I'm looking for "bad series", we have 3 possible arrangements to a "bad series".

Now I handle the "bad series" as 1 item, so the total number of items I have in the deck is 21. with 21! ways to order them.

The probability to find the "bad series" in the pulling order is:

$$P=\frac{21!}{30!}$$

but the number of "bad series" we have is 3, so the total probability is: $$P_{total}=\frac{3\times 21!}{30!}$$

so the probability that each of those "bad series" will not show up is:

$$P_{final}=1-\frac{3\times 21!}{30!}$$

Is my way of thinking correct? are there any other ways to approach those types of problems?

knight5478
  • 347

1 Answers1

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The first part is correct but you forget that there are sequences where two or all three of your series show.

Therefore you need to account for $3\times\frac{12!}{30!}$ for the ones where there are two of your series and $\frac{3!}{30!}$ for when there are all three. In total you get $$P_{final} = 1-\frac{3\times21!}{30!}+\frac{3\times12!}{30!}-\frac{3!}{30!}$$

munichmath
  • 96