How to choose k in Poisson distribution? 
Problem: In a factory, the probability of a screw being defective is p
  = 0.015. What is the probability that a box of 100 screws does not contain a defective one?

One way to answer this problem is to use the Poisson formula the following way:
a) $P (\textrm{There are no defective screws in a box of 100}) = e^{-\lambda} \cdot \frac{\lambda^k}{k!}$,
   where $\lambda = n\cdot p = 100 \cdot  0.015 = 1.5$,
k = 0.   
Hence, $e^{-1.5} \cdot \frac{1.5^0}{k!} = e^{-1.5} \cdot  (1 / 1) = 0.2231$, which is the correct answer.
I am confused, however, why this problem cannot be calculated in the reverse way, by focusing on the probability that all 100 screws are good (which should be the same that none of them are defective). This is what I mean:
b) $P (\textrm{All screws are good in a box of 100}) =e^{-\lambda} \cdot \frac{\lambda^{k}}{k!}$,
where $\lambda = n\cdot p = 100 \cdot  (1-0.015) = 100 \cdot 0.985 = 98.5,  
      k = 100.$
Hence, $e^{-98.5} \cdot  \frac{98.5^{100}}{ 100!} = \frac{3.678\cdot 10^{156}}{100!} = 0.039$, which is clearly not the same as a). 
Is there any stipulation in the Poisson formula whether $k$ stands for success or failure? If yes, then how does one know?
(The exercise is from Feller: An Introduction to probability theory and its applications, Vol.1, p- 155/d.)


"The Poisson distribution can be applied to systems with a large
  number of possible events, each of which is rare"

wiki, Poisson distribution
This is not the case if $p=0.98$. Basically the random variables is binomial distributed. This variable can be approximated by the Poisson distribution if $n$ is large and $p$ is small.
If we use the binomial distribution we obtain $P(X=100)=0.985^{100}\approx 22\%$
