Probability toss a sequence of coins

Suppose you toss a sequence of coins. The iith of which comes up heads with probability pi,∑∞i=1pi<∞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? =================      I feel as though the probability is 00, but I don't have a proof off the top of my head. – Brian Tung 2 days ago ================= 2 Answers 2 ================= Let AiA_i be the event that coin ii lands on heads. Then: ∞∑i=1P(Ai)=∞∑i=1pi<∞.\sum_{i=1}^\infty P(A_i)=\sum_{i=1}^\infty p_i<\infty. By the Borel Cantelli lemma, this implies the probability of AiA_i occurring infinitely often is 0. It follows that the probability of infinitely many heads is 0. We shall assume that all the pip_i be strictly less than 11, since if some coins have p=1p=1 we can alltogether remove them from the considered pool. Therefore: 0⩽pi<1|∀i 0 \leqslant p_{\,i} < 1\quad \left| {\;\forall i} \right. Now if they are less than 11, 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. 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} \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.