Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer
Can you help me with this problem: Let X1 ,X2 , . . . ,Xk-1 have a multinomial distribution. Find the mgf of X2,X3,...,Xk-1. – user48495 Nov 7 '12 at 14:49
@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… – raindrop Jun 14 '13 at 1:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.