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I have $100$ balls and on each ball is labelled a number between $1$ and $100$. No two balls can have the same number, so the interval $1\ldots 100$ is represented by the balls.

I now pick 20 balls at random, one by one without putting them back at any point. I am trying to find the probably of getting 4 certain numbers. The sequence of these 4 numbers does not matter.

Here is my solution: The first time I pick up a ball the probably is $1/100$. The next time it is $1/99$, etc.. So the total probably of getting the four desired numbers is $1/(100\cdot 99\cdot 98\cdot 97)$.

Is this correct?

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You have computed the probability that you get your target numbers, in the first $4$ choices, in a certain specified order. That is much smaller than the required probability.

There are $\binom{100}{20}$ equally likely ways to choose $20$ balls.

We find the number of ways to choose $20$ balls that include our $4$ target balls. So we must choose $16$ balls from the remaining $96$ to keep our $4$ target balls company. This can be done in $\binom{96}{16}$ ways.

Thus the required probability is $\dfrac{\binom{96}{16}}{\binom{100}{20}}$. This expression can be greatly simplified.

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  • $\begingroup$ Thanks, but the probablity of picking the $4$ does not enter in the equation? $\endgroup$ – BillyJean Jan 20 '14 at 19:01
  • $\begingroup$ Well, on top we could write $\binom{4}{4}\binom{96}{16}$ instead of $\binom{96}{6}$, and maybe I should have, but since $\binom{4}{4}=1$ it makes no difference. $\endgroup$ – André Nicolas Jan 20 '14 at 19:03
  • $\begingroup$ But the balls are not supposed to be returned. Is $\binom{100}{20}$ correct? $\endgroup$ – Spock Jan 20 '14 at 19:07
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    $\begingroup$ Yes. It is the number of ways of choosing a "hand" of $20$ cards from a collection of $100$ distinct cards. $\endgroup$ – André Nicolas Jan 20 '14 at 19:10
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Your solution is not correct. But, here's a hint on a different way to proceed:

Hint: You want to count all permutations (of length $20$) of distinct elements in $\{1,2,\ldots,100\}$, such that four fixed numbers (say $1,2,3,4$) are all contained in the permutation.

To do this: pick the $16$ numbers OTHER than the ones you are interested in which will be included; then, pick ANY arrangement of the set of $20$ numbers consisting of the four you want and the new sixteen.

This is the total number of sequences with the desired property; divide by the total number of sequences to get the desired result.

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  • $\begingroup$ But I get to pick 20 balls, how does that information enter? $\endgroup$ – BillyJean Jan 20 '14 at 18:49
  • $\begingroup$ Whoops, sorry; I missed that detail. One moment. $\endgroup$ – Nick Peterson Jan 20 '14 at 18:50
  • $\begingroup$ @BillyJean Take a look at it now. $\endgroup$ – Nick Peterson Jan 20 '14 at 18:53

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