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.
 A: 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!
A: Yes, you are on the correct track, but this might be a bit easier to reason about.


*

*Change the problem into a Bernoulli random variable problem by considering this event: a biscuit has more than 7 chips or not. Let $$p = P(X > 7) = \sum_{i = 8}^{\infty}\frac{\lambda^ie^{-\lambda}}{i!},$$ where $\lambda = 5$.

*Then, our problem becomes what is the probability that out of 20 biased coin flips with probability $p$ of success, at least one will succeed?


This would be found using the reasoning: what is the probability that NONE of them succeed? $(1 - p)^{20}$. We then subtract this from one to get our desired value: $$1 - (1 - p)^{20}.$$ 
$p$ can be calculated using a computer, by the way. It is simply a matter of routine calculations.
A: Hint: please note that the box has 20 biscuits and they are independent, and the number of chocolate chips in (one*) biscuit follows a Poisson distribution with $\lambda=5$, so first you need to compute the probability that one biscuit has more than 7 chocolate chips and then find the desired probability.
