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At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of 5 phone calls during 10-minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is
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$\frac{6}{e^5}$
$\frac{6}{e^5}$
$P(X=r)=\frac{e^{-m} m^r}{r !}$
$P(X \leq 1)=P(X=0)+P(X=1)$
$=e^{-5}+5 \times e^{-5}=\frac{6}{e^5}$
$P(X \leq 1)=P(X=0)+P(X=1)$
$=e^{-5}+5 \times e^{-5}=\frac{6}{e^5}$
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