Easy way to compute $Pr[\sum_{i=1}^t X_i \geq z]$ We have a set of $t$ independent random variables $X_i \sim \mathrm{Bin}(n_i, p_i)$.
We know that $$\mathrm{Pr}[X_i \geq z] = \sum_{j=z}^{\infty} { n_i \choose j } p_i^j (1-p_i)^{n_i -j}.$$
But is there an easy way to compute:
$$\mathrm{Pr}\left[\sum_{i=1}^t X_i \geq z\right]?$$
MY IDEAS:
This should have to do something with convolution, but I am not sure.
Is it easier to compute
$$\mathrm{Pr}\left[\sum_{i=1}^t X_i \leq z \right]?$$
What I thought of is maybe:
$$\mathrm{Pr}\left[\sum_{i=1}^t X_i \leq z \right]=\sum_{j=1}^{z} \mathrm{Pr}\left[\sum_{i=1}^t X_i = z \right]$$
but this seems to be quite hard with $t$ random variables. 
Would appreciate any hint and if you don't write an answer I am interested in whether it is too difficult or too easy?! thank you.. 

I'll give you a general idea. If you want the details, look up an article by Tomas Woersch (here).
So you have 
$$
S_m(n) = \sum_{k=0}^{m} \binom{n}{k}x^k = 1 + \binom{n}{1}x +\ldots + \binom{n}{m}x^m\\
\frac{S_m(n)}{\binom{n}{m}x^m} = 1 + \ldots + \frac{\binom{n}{1}x}{\binom{n}{m}x^m} + \frac{1}{\binom{n}{m}x^m} = 1 + \ldots a_{m-1}x^{1-m} +a_m x^{-m}
$$
Now you need to do the following: obtain the ration of binomial coefficients I denoted with $a_k$ and then approximate then using Stirling approach. Use the fact that 
$$(1 - o(1))\sqrt{2 \pi n} (\frac{n}{e})^n \leq n! \leq (1+o(1))\sqrt{2 \pi n} (\frac{n}{e})^n$$
