Probability toss a sequence of coins Suppose you toss a sequence of coins. The $i$th of which comes up heads with probability $p_i, \sum_{i=1}^\infty p_i < \infty$. Calculate the probability of the event that infinitely many heads occur.
This problem focused on the Bernoulli test scheme with n tending to infinity? Can I use Poisson distribution to solve this problem? Or must I use Stirling's formula and lage number law?
 A: Let $A_i$ be the event that coin $i$ lands on heads. Then:
$$\sum_{i=1}^\infty P(A_i)=\sum_{i=1}^\infty p_i<\infty.$$
By the Borel Cantelli lemma, this implies the probability of $A_i$ occurring infinitely often is 0. It follows that the probability of infinitely many heads is 0. 
A: We shall assume that all the $p_i$ be strictly less than $1$, since if some coins have $p=1$ we can alltogether remove them from the considered pool.
Therefore:
$$
0 \leqslant p_{\,i}  < 1\quad \left| {\;\forall i} \right.
$$
Now if they are less than $1$, we can find a max for their value, and fix
$$
p_{\,i}  \leqslant 1 - \frac{1}
{q} = \frac{1}
{{\frac{q}
{{q - 1}}}} = \frac{1}
{{1 + \frac{1}
{q}}} = \frac{1}
{{1 + \varepsilon }}\quad \left| \begin{gathered}
  \;1 < q \hfill \\
  \;0 < \varepsilon  \hfill \\ 
\end{gathered}  \right.
$$
Then
$$
\begin{gathered}
  0 \leqslant \prod\limits_{i = 1}^n {p_{\,i} }  \leqslant \frac{1}
{{\left( {1 + \varepsilon } \right)^{\,n} }} \hfill \\
  0 \leqslant \mathop {\lim }\limits_{n\, \to \,\infty } \prod\limits_{i = 1}^n {p_{\,i} }  \leqslant 0 \hfill \\ 
\end{gathered} 
$$
as it shoud clearly be.
