If you take k balls out of a bunch of N balls, the probability of
having selected k balls is always one(1.0). It takes no calculation,
because after you sum that expression up, it add up to one by
definition/law of probability.
On Jan 23, 5:12 pm, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote:
> I need to compute what is the probability of selecting k balls from N
> balls where ball j can be picked with prob p_j efficiently.
>
> Prob (k balls are picked) = \sum_{i_1}^N \sum_{i_2, i_2 not equal to
> i_1}^N......\sum_{i_k=1, i_k not equal to any of i_1,i_2,..,i_{k-1}}^N
> p_{i_1} p_{i_2}...p_{i_k} \Prod_{j not equal to i_1, i_2, ..,i_k} (1-
> p_j)
>
> how can I compute Prob (k) efficiently, it seems there are N choose K
> terms inside and for me N = 350 and K=10 ?
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