The Probability of Making a Mistake Given Discrete Events This feels like grade-school level probability but I'm having a bit of trouble thinking about it :( 
Broadly speaking: I'm predicting $n$ independent Bernoulli events, what's the probability of me making at most $k$ ($k < n$) number of mistakes? 
I try to formulate the problem into a more formal setting:
Given the Bernoulli distribution of a set of independent events $\mathcal{A} = \{e_1, e_2, e_3, ..., e_n \}$, the probability of me not making a mistake (or success): $p(e_i)$ is known and fixed. (for an example: $\mathcal{A} = \{e_1, e_2, e_3\}$, and $p(e_1)=0.3, p(e_2)=0.6, p(e_3)=0.7$, each one means how likely am I going to succeed at event $i$).
I want to use these statistics (but possibly I can collect more...if needed), to compute the following:


*

*What is the probability of me with complete success (making zero mistake)? (I'm thinking it's $1 - \prod_i (1-p(e_i))$)

*What is the probability of me making at least one mistake? (regardless of which one)

*Is there a formula equation for this $\forall n \forall k$?


Any help would be appreciated!! 
 A: Just a heads up:
For $1)$, the probability is $\prod_{i=1}^{n}p(e_i)$ since this amount to every trial being a success using independence of the trials. Your expression shows the probability that we have at least one success since $\prod_{i=1}^{n}(1-p(e_i))$ is the probability of no successes.
For $2)$ we have that the probability is $1-\prod_{i=1}^{n}p(e_i)$ since the complement of the event $\{\text{no mistakes}\}$ is the event $\{\text{at least one mistake}\}$
As far as I can tell, a general formula could be made but it would be disgusting:
For example, for $1$ success, you can have the formula: $$\mathbb{P}(\text{1 success})=\sum_{j=1}^{n}p(e_j)\prod_{i=1,i\neq j}^{n}(1-p(e_i))$$
What this basically says is that we first choose the success and then make all the other trials fail.
For more successes it would be similar, though you'll have to use binomial coefficients so that you can choose the successes out of all trials.
Finally, if all of the trials are identical and independent Bernoulli, then you can have a look at the Binomial distribution since it is a sum of the said Bernoulli variables(they have to be independent and identically distributed to work).
A: *

*This is rather $\prod p(e_i)$ given your definition of $p(e_i)$. Independence turns "and"s into products.

*This is $1-$ whatever the answer to 1. was.

*The probability of getting it right $k$ times is the following sum:
$$\sum_{I\subset \{1, \ldots, n\}\\\ \ \ \ \ |I|=k}\left(\prod_{i\in I}p(e_i)\right)\left(\prod_{j\not \in I}1-p(e_j)\right)$$
