A question regarding the Poisson distribution The number of chocolate chips in a biscuit follows a Poisson distribution with and average of $5$ chocolate chips per biscuit. Assume that the numbers of chocolate chips in different biscuits are independent of each other. What is the probability that at least one biscuit in a box of $20$ has more than $7$ chocolate chips? Let $X$ be the number of chocolate chips in a biscuit. We know that $\lambda = E[X]=5$. Then the probability that each biscuit has more than $7$ chocolate chips is $$\Pr(X \gt 7) = 1 - \Pr(X \le 6) =p_1,$$ where $p_1$ is a value to be found. Let $Y$ be the number of biscuits in a box of $20$ that has at more than $7$ chocolate chips. Then the probability that at least one biscuit in a box of $20$ has more than $7$ chocolate chips is $$\Pr(Y \le 1) = \Pr(Y=0) + \Pr(Y=1) = \frac{{p_1}^0 e^{-p_1}}{0!} + \frac{{p_1}^1 e^{-p_1}}{1!}=e^{-p_1}(1+p_1).$$ Is this correct? I appreciate your help.

The probability that at least one has more than $7$ is, in the notation of your post, $$1 -\left(\Pr(X\le 7)\right)^{20}.$$

This because the event "at least one has more than $7$" is the complement of the event "all $20$ have $\le 7$." By independence, the probability that $20$ cookies in a row have $\le 7$ is $\left(\Pr(X\le 7)\right)^{20}$.

**Remark:** In the post, you were looking at $\Pr(X\le 6)$. That's not quite the relevant probability, since the problem says _more than_ $7$. Similarly, in the attempted calculation of the probability of at least one, the wrong index was being talked about. The probability of at least one is $1$ minus the probability of _none_.

In the answer, I assumed you know how to find $\Pr(X\le 7)$. We have $$\Pr(X\le 7)=\sum_{k=0}^7 e^{-\lambda}\frac{\lambda^k}{k!},$$ where $\lambda=5$. A somewhat tedious calculation!