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.
poisson distribution of chocolate
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statistics
1 Answers
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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.
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0@user48495 "We know that the MGF of a multinomial distribution is: $M(t_1, t_2,...t_k-1) = (p_1*e^t_1 + .... + p_k+1*e^t_k+1 + p_k)^n$ hence, the MGF of $X_2$, $X_3$, .... $X_k-1$ is $M(0, t_2,...t_k-1) = (p_1 + .... + p_k+1*e^t_k+1 + p_k)^n$" -adapted from rad at http://www.actuarialoutpost.com/actuarial_discussion_forum/archive/index.php/t-121247.html – 2013-06-14