Bernouilli trial? not sure Suppose a box, with 100 balls where 3 of them are black. No replacement.
The odds of not picking a black ball after 10 balls are prod ((97-n)/(100-n)) from n=0 to n=9;
the odds of picking 1 black ball after 10 trials are 3/91;
the odds of picking 2 black balls after 10 trials are ((3/92)*(2/91)).
What are the odds of picking less than 2 black balls after take n balls?
Peter, thanks it's the hypergeometric distribution
 A: I will assume you wish to find the probabilities rather than the odds.  
The probability that exactly $k$ black balls are selected when $10$ balls are selected without replacement is 
$$\frac{\dbinom{3}{k}\dbinom{97}{10 - k}}{\dbinom{100}{10}}$$
Hence, the probability that no black balls are selected when $10$ balls are selected without replacement is 

 $$\frac{\dbinom{3}{0}\dbinom{97}{10}}{\dbinom{100}{10}}$$

The probability that exactly one black ball is selected when $10$ balls are selected without replacement is 

 $$\frac{\dbinom{3}{1}\dbinom{97}{9}}{\dbinom{100}{10}}$$

The probability that exactly two blacks balls are selected when $10$ balls are selected without replacement is 

 $$\frac{\dbinom{3}{2}\dbinom{97}{8}}{\dbinom{100}{10}}$$

The probability that exactly three black balls are selected when $10$ balls are selected without replacement is 

 $$\frac{\dbinom{3}{3}\dbinom{97}{0}}{\dbinom{100}{10}}$$

Since these cases are mutually exclusive and exhaustive, their probabilities should add to $1$, as you can verify.
Finally, the probability that less than two black balls are selected when $10$ balls are selected without replacement is found by adding the results for the probabilities that no black balls are selected and exactly one black ball is selected.  
