<s>[INST] How to calculate probability of $5$ in a row with at least one rare vs $5$ in a row without a single rare? There are $36$ possible outcomes. Event $A$ (later - "rare group") happens in $11$ outcomes. Event $B$ (normal group) in $25$ outcomes ($36-11$). Events $A$ and $B$ cannot occur at the same time - only one of them is possible. 
We make $5$ Bernoulli trials and observe the composite result of all $5$ trials. After any event happens, number of "positive" outcomes for that event decreases by one (if event $A$ happened, next trial there are $35$ (not $36$) outcomes in general for both $A$ and $B$, but now only $10$ for $A$ (that is minus one), still $25$ for $B$ (nothing changed for B positive outcomes)). I need to calculate and compare probability that in $5$ trials 


*

*Event $A$ never appears,

*Event $A$ appears at least once.


I have an idea how to calculate, but I get a result more than $1$, which looks bad. So I am wrong somewhere.
The question is about $5$ Bernoulli trials (tossing a single "die" with $36$ sides). Each next trial (appearance of next number) is a conditional probability, because trial $2$ is equivalent of tossing a single "die" with $35$ sides ($36$ - what appeared in trial 1). So in trial 1 probability to get a number within rare group is $11/36$, probability to get a number not in rare group is $1-11/36$. In trial 2: probability to get a number within rare group is $11/35$ (UNLESS THAT NUMBER APPEARED IN TRIAL 1! Then it is $10/35$), probability to get a number not in rare group is $1 - 11/36$. Finally, probability that all $5$ numbers (results of all $5$ trials) are not from rare group is just a product of respective probabilities of trials 1-5: 
$$\left(1 - \frac{11}{36}\right)\left(1 - \frac{11}{35}\right)\left(1 - \frac{11}{34}\right)\left(1 - \frac{11}{33}\right)\left(1 - \frac{11}{32}\right) = \boldsymbol{14\%}$$ 
Same as $$\frac{\dbinom{25}{5}}{\dbinom{36}{5}}$$ 
$C = \binom{25}{5}$ ways to choose $5$ without replacement out of $25$, $\binom{36}{5}$ - total combinations.
So, the opposite event (probability that a ball from rare group would appear at least once) is $100\% - 14\% = 86\%$. Right? But why...
On the other hand, probability that a ball from rare group would appear at least once in $5$ trials is a sum of respective probabilities: $$\frac{11}{36} + \frac{11}{35} + \frac{11}{34} + \frac{11}{33} + \frac{11}{32} = 1.62 = 162\%$$ 
(no adjustments in case of appearing of rare ball in trials 1-4, but even all five rare balls is huge (rough estimation): 
$$\frac{11}{36} + \frac{10}{35} + \frac{9}{34} + \frac{8}{33} + \frac{7}{32} = 1.31 = 131\%$$
 [/INST] Since the selections are made without replacement, this is actually a hypergeometric distribution.  
You correctly calculated that the number of ways no members of the rare group are selected in five trials is 
$$\frac{\dbinom{25}{5}}{\dbinom{36}{5}}$$
Therefore, the probability that at least one member of the rare group is selected in five trials is 
$$1 - \frac{\dbinom{25}{5}}{\dbinom{36}{5}}$$
The number of ways of selecting exactly $k$ members of the rare group and $5 - k$ members of the normal group in five trials is 
$$\binom{11}{k}\binom{25}{5 - k}$$
Hence, another way of calculating the probability that at least one member of the rare group is selected is to add the probabilities that from $1$ to $5$ members of the rare group are selected, which is 
$$\frac{\dbinom{11}{1}\dbinom{25}{4} + \dbinom{11}{2}\dbinom{25}{3} + \dbinom{11}{3}\dbinom{25}{2} + \dbinom{11}{4}\dbinom{25}{1} + \dbinom{11}{5}\dbinom{25}{0}}{\dbinom{36}{5}}$$
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