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Suppose the probability for getting head is $p_i$ for $i$th coin.

You can easily (and economically) compute the probabilities of exactly $k$ heads using the recursive relation -

$H_{n,k}=p_nH_{n-1,k-1}+(1-p_n)H_{n-1,k}$

Explanation follows.

Let $H_{n,k}$ be the probability of getting exactly $k$ heads using the first $n$ coins. For answering the type of questions you want to solve, all you need is a list of $H_{n,k}$'s.

Note that $H_{n,k}=\sum_{\left[e_i\in\{0,1\},\sum e_i=k\right]}p_i^{e_i}(1-p_i)^{1-e_i}$H_{n,k}=\sum_{\left[over\ e_i's\in\{0,1\},\sum_i^n e_i=k\right]}\prod_{i=1}^n p_i^{e_i}(1-p_i)^{1-e_i}$The sum (as you mentioned) contains$n\choose k$entries. However note that$H_{n,k}=p_nH_{n-1,k-1}+(1-p_n)H_{n-1,k}$So you can recursively build up the$H_{n,k}$'s which should be simple since there are only a few of them. (To be precise, for$N$coins, there are$N(N+3)/2$many of$H_{n,k}$'s since$n\in \{1,...,N\}$and$k\in \{0,...,n\}$). As a base for the recursive relation, you can use the following (obvious) identities. •$H_{n,k}=0$for$k\gt n$•$H_{n,0}=\prod_{i=1}^n(1-p_i)$•$H_{n,n}=\prod_{i=1}^np_i$5 added 122 characters in body; added 5 characters in body Suppose the probability for getting head is$p_i$for$i$th coin. You can easily (and economically) compute the probabilities of exactly$k$heads using the recursive relation -$H_{n,k}=p_nH_{n-1,k-1}+(1-p_n)H_{n-1,k}$Explanation follows. Let$H_{n,k}$be the probability of getting exactly$k$heads using the first$n$coins. For answering the type of questions you want to solve, all you need is a list of$H_{n,k}$'s. Then Note that$H_{n,k}=\sum_{\left[e_i\in\{0,1\},\sum e_i=k\right]}p_i^{e_i}(1-p_i)^{1-e_i}$The sum (as you mentioned) contains$n\choose k$entries. However note that$H_{n,k}=p_nH_{n-1,k-1}+(1-p_n)H_{n-1,k}$So you can recursively build up the$H_{n,k}$'s which should be simple since there are only$nk$many a few of them. (To be precise, for$N$coins, there are$N(N+3)/2$many of$H_{n,k}$'s since$n\in \{1,...,N\}$and$k\in \{0,...,n\}$). As a base for the recursive relation, you can use the following (obvious) identities. •$H_{n,k}=0$for$k\gt n$•$H_{n,0}=\prod_{i=1}^n(1-p_i)$•$H_{n,n}=\prod_{i=1}^np_i$4 added 128 characters in body Suppose the probability for getting head is$p_i$for$i$th coin. You can easily (and economically) compute the probabilities of exactly$k$heads using the recursive relation -$H_{n,k}=p_nH_{n-1,k-1}+(1-p_n)H_{n-1,k}$Explanation follows. Let$H_{n,k}$be the probability of getting exactly$k$heads using the first$n$coins. For answering the type of questions you want to solve, all you need is a list of$H_{n,k}$'s. Then$H_{n,k}=\sum_{\left[e_i\in\{0,1\},\sum e_i=k\right]}p_i^{e_i}(1-p_i)^{1-e_i}$The sum (as you mentioned) contains$n\choose k$entries. However note that$H_{n,k}=p_nH_{n-1,k-1}+(1-p_n)H_{n-1,k}$So you can recursively build up the$H_{n,k}$'s which should be simple since there are only$nk$many of them. As a base for the recursive relation, you can use the following (obvious) identities. •$H_{n,k}=0$for$k\gt n$•$H_{n,0}=\prod_{i=1}^n(1-p_i)$•$H_{n,n}=\prod_{i=1}^np_i\$
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