The number of accidents per week in a town follows Poisson distribution with mean 3. If the probability that there are three accidents in two weeks time is \(ke^{-6}\), then the value of k is
$$\begin{aligned} \text{Weekly Mean } (\lambda_1) &= 3 \\ \text{Two-Week Mean } (\lambda_2) &= 3 \times 2 = 6 \\ \text{Poisson PMF: } P(X=x) &= \frac{e^{-\lambda} \cdot \lambda^x}{x!} \\ P(X=3) &= \frac{e^{-6} \cdot 6^3}{3!} \\ &= \frac{e^{-6} \cdot 216}{6} \\ &= 36e^{-6} \\ ke^{-6} &= 36e^{-6} \\ k &= 36 \end{aligned}$$
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