Solving a Poisson Distribution The number of contaminants in samples of one cubic centimeter of polluted water has a Poisson
distribution with a mean of 1.
a. What is the probability a sample will contain some (one or more) contaminants?
b. If four samples are independently selected from this water, find the probability that at least
one sample will contain some (one or more) contaminants?
c.If 100 samples are selected instead, what is approximately the probability of seeing at least
60 samples with a contaminant? (Use the Normal table provided for your calculations)
This is a practice final exam, however our teacher does not respond to emails and we do not have the solutions. For a. I got 1 - the probability of zero samples getting contaminated and I am fairly certain that it is correct, however I am stumped on b. and c. Any help would be greatly appreciated.
 A: Yes, for the first the answer is $1-e^{-1}$, for the reason you described. 
For the second, it is $1$ minus the probability they are all clean. The probability that any given one is clean is $e^{-1}$. So the probability they are all clean is $(e^{-1})^4$, and therefore the probability of at least one dirty is $1-e^{-4}$. 
For the third, let $p$ be the probability that an individual sample has $1$ or more contaminant. We have $p=1-e^{-1}$, but we will use $p$.
Let $X$ be the number of dirty in a sample of $100$. Then $X$ has binomial distribution $n=100$, $p=p$.
The probability $60$ or more are dirty is then
$$\sum_{k=60}^{100}p^k (1-p)^{100-k}.$$
The distribution of the random variable $X$ is well approximated by the normal with mean $\mu=100p$, variance $\sigma^2=100p(1-p)$. Compute these two numbers to reasonable accuracy. 
Now compute the probability that a normal $Y$ with these parameters is $\ge 60$. If in your work you use the continuity correction, it would be better to find $\Pr(Y\gt 59.5$. I assume you know how to calculate probabilities that involve the normal.
