Poisson Approximation to Binomial I have a Poisson approximation to binomial question, posted below. I'm not too sure if I'm using the proper formula: 
$$P(x) = e^{-np}(np)^x/x!$$ .     
Of anyone can tell me if I'm doing this right, that would be great. 
Again these are just practice questions not homework.

Q: A salesperson has found that the probability of making a sale on a
  particular product manufactured by him or her company is .05. If the salesperson
  contacts 140 potential customers, what is the probability he or she will sell at least
  2 of these products? Use and justify Poisson approximation to Binomial.

What I'm doing:
$$P(x) = e^{-140 (.05)}(140*.05)^2/2!$$ .   
 A: According to Wikipedia: Poisson approximation, the use of this approximation seems ok as n=140 and np=7

Poisson approximation The binomial distribution converges towards the
  Poisson distribution as the number of trials goes to infinity while
  the product np remains fixed. Therefore the Poisson distribution with
  parameter λ = np can be used as an approximation to B(n, p) of the
  binomial distribution if n is sufficiently large and p is sufficiently
  small. According to two rules of thumb, this approximation is good if
  n ≥ 20 and p ≤ 0.05, or if n ≥ 100 and np ≤ 10.[9]

From wikipedia, it is clear your P(x) is a mass function where giving the probability of a particular value of x, not a cumulative distribution function giving the probability of a value below or equal to x.
To find the probability for x (the number of sales) to be at least 2, you can calculate either:
Prob{x>=2} = Sum(P(x) for integer x>=2) which is an infinitely long sum, or
Prob{x>=2} = 1 - Prob(x<2) = 1 - P(0) - P(1)

I would imagine the second method to be easier, yielding:
Prob{x>=2} = 1 - exp(-7) - 7 exp(-7) = 1 - 8 exp(-7) = 0.9927
