poisson distribution of chocolate

Question

Let the number of chocolate drops in a certain type of cookie have a Poisson distribution. We want the probability that a cookie of this type contains at least two chocolate drops to be greater than 0.99. Find the smallest value of the mean that the distribution can take.

Answer 1

Let's say the number $D$ of drops has a distribution $D \sim \mathrm{Poisson}(\lambda)$. Then \begin{align*} P(D \ge 2) &= 1 - P(D < 2)\\ &= 1 - P(D = 0) - P(D = 1)\\ &= 1 - \exp(-\lambda)\cdot (1 + \lambda) \end{align*} So $P(D \ge 2) \ge 0.99$ iff $\exp(-\lambda)(1 + \lambda) \le \frac 1{100}$. As $\lambda \mapsto \exp(-\lambda)(1+ \lambda)$ has a negative derivative on $(0,\infty)$, it is strictly decreasing. So there is a unique $\lambda_0$ with $\exp(-\lambda_0)(1+\lambda_0) = \frac 1{100}$ (there is no closed form for $\lambda_0$ in terms of elementary functions, Wolfram|Alpha tells us it is approximately 6.6384). This is the minimal mean you looked for.