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A telephone switchboard can make \\(\\var{m}\\) connections per second on average. 

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The Poisson distribution \\(P(X=k)=\\frac{e^{-\\lambda}\\lambda^k}{k!}\\)

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In this problem \\(\\lambda=\\var{m}\\)

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P(at most \\(\\var{n}\\) connections) = \\(P(X\\le\\var{n})\\)

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\\(P(X\\le\\var{n})=P(X=0)+P(X=1)+...P(X=\\var{n})\\)

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\\(P(X=0)=\\var{p_0}\\)

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\\(P(X=1)=\\var{p_1}\\)

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\\(P(X=2)=\\var{p_2}\\)

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\\(P(X=3)=\\var{p_3}\\)

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\\(P(X=4)=\\var{p_4}\\)

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\\(P(X=5)=\\var{p_5}\\)

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\\(P(X\\le\\var{n})=\\var{p}\\)

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Calculate, correct to three decimal places, the probability that the switchboard will make at most \\(\\var{n}\\) connections in a particular minute.    [[0]]

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