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The question to be more clear is:

Total no. of coins: 100

Rare coins in the set: 41

What is the chance of picking up 30 rare coins continuously?

My attempts at solving it:

First probability = $\frac{41}{100}$; Second probability = $\frac{40}{99}$; Third probability = $\frac{39}{98}$; ......; Thirtieth probability = $\frac{11}{70}$

Chances of getting 30 rare coins in a row = $\frac{41*40*39*........*11}{100*99*98*......*70} = \frac{41!}{11!} * \frac{70!}{100!}$

Am I on the right path to solving this question? Or is there an easier way?

Thanks.

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    $\begingroup$ Looks good. As a minor variant, you could remark that there are $\binom {100}{30}$ ways to choose $30$ coins from the pile, and $\binom {41}{30}$ ways to choose them from the rare subset, then just take the ratio. Of course that just returns the same fractions you already discovered. $\endgroup$ – lulu Dec 12 '16 at 2:06
  • $\begingroup$ And to add to what @lulu said, this is called a combination and is similar to what you may have learned about the binomial coefficient. $\endgroup$ – rb612 Dec 12 '16 at 2:08
  • $\begingroup$ @rb612 how would I find an exact answer from this ratio? My calculator overflows and cannot calculate it, do I need to use a computer or is there a way I do not know of? $\endgroup$ – Sarhad Salam Dec 12 '16 at 2:16
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    $\begingroup$ @SarhadSalam I would use Wolfram Alpha $\endgroup$ – rb612 Dec 12 '16 at 2:38
  • $\begingroup$ One way to compute a numerical result is to take logarithms, i.e. compute the logarithm of the answer. You can avoid computing the large factorials like 100! by using Stirling's approximation to n! in order to find log(100!) approximately. $\endgroup$ – awkward Dec 12 '16 at 11:50
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One other way to solve this.

$\frac{C(41,30)}{C(100,30)}$

$= \frac{\frac{41!}{30!*11!}}{\frac{100!}{30!*70!}}$

$= \frac{41!}{11!} * \frac{70!}{100!}$

And when calculations are too long you don't need to solve it. It becomes answer.

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