Poisson distribution help... please?!?

Question

These are some questions from my study guide. I have the answers, but I don't understand it. =(

The Securities and Exchange Commission has determined that the number of companies listed in NYSE declaring bankruptcy is approximately a Poisson distribution with a mean of 2.6 per month.

26. Find the probability that more than 1 bankruptcy occur next month.
A) .1931
B) .9257
C) .7326
D) .4816
E) .2674

Answer: C

27. Find the probability that no more than 1 bankruptcy occur next month.
A) .1931
B) .9257
C) .7326
D) .4816
E) .2674

Answer: E

Answer 1

For the Poisson distribution, the probability of k successes with mean λ is given by:

P( k | λ) = {[e^(-λ)] * (λ^k)} / k!

Given λ = 2.6

27) Find the probability that no more than 1 bankruptcy occur next month.

P(k = 0 or 1 | 2.6) = P(0 | 2.6) + P(1 | 2.6)

= 0.074273578 + 0.193111303 ≈ 0.2674

The answer is E.
_______________

26) Find the probability that more than 1 bankruptcy occur next month.

P(k > 1 | 2.6) = 1 - P(0 | 2.6) - P(1 | 2.6)

= 1 - 0.2674 = 0.7326

The answer is C.

Answer 2

You may be aware that if you add the probability of the first to the second Q, they add to 1. (they are mutually exclusive and form the whole of probabilities for 1 bankruptcy).
So even if you don't know anything else about the distribution, you could conclude that C& E would need to be your answers in one order or another (C then E, or E then C) as they are the only two answers that add to one (if you are asked for the probability that either something will happen or it will not, that is a probability of 1 (a certainty), as obviously something either will or will not happen, there is no middle ground.
The next part requires you to consider wihich of the answers would be closer to 1. Do you think that it would be more or less likely that if the mean is 2.6, that there would be more or not more than 1 bankruptcy. Logically, if the mean is 2.6, it is more likely that more than 1 bankruptcy will occur, so you would have to choose C for Q26.
Another way to think of it is that it would be equally likely (probability 0.5) that either more than or not more than the mean (2.6 in this case would occur), so the probality that more than one 1 occurence would need to be less than the probability of not less than 2.6. (<0.5)

Answer 3

According Poisson's formula for a random variable X with Poisson distribution with mean 2.6 we have:
P(X = k) = 2.6^k*e^(-2.6)/k!, where k = 0,1,2,3,....; then

27. P(X ≤ 1) = P((X=0) or (X=1)) = P(X=0) + (X=1) =
= 2.6^0*e^(-2.6)/0! + 2.6^1*e^(-2.6)/1! ≈ 0.27 /answer E/;

26. P(X > 1) = P(not (X ≤ 1)) = 1 - P(X ≤ 1) ≈ 1 - 0.27 ≈ 0.73 /answer C/

In the Poisson distribution table I have there is mean 2.5 but 2.6 is missing, so I had to approximate.