3 Laws of Miracles and the Poisson distribution

ok i've to admit, i've gone crazy over this thing.

Let X be the discrete random variable denoting the number of miracles that happen over a stipulated, fixed time t, over infinite trials. The probability of a miracle happening is p.

As miracles are rare, n is large, n > 50 (n = times that miracles could happen in the time frame) and p < 0.01, approximate to a Poisson distribution.

X ~ Po (np) approximately

With approximation,

P (X < x ) = [e^-(np) ][(np)^-x] / [x!]

lets say probability of miracle happening is 10^-9, as in, a one in a billionth chance.
assume that every day for a period of 21 years, a miracle could happen.

therefore, n = 21 x 365 = 7665; np = 7.665 x 10^-6

a) what is the most likely number of miracles that will happen during any given day during this 21 years?

Using recurrence formula,

[P (X < x + 1)] / [P (X < x)] = lambda / (x + 1) = np /(x + 1)

Mode: np/(x+1) > 1

7.665 x 10^-6 > x + 1
x < 1

Since x is an integer, x = 0
hence mode = 0

hence, most probably, there will most likely be no miracles happening for any given day in this 21 years. =(

b) what is the probability of at least one miracle happening on any given day in this 21 years?

P (X < 1) = 1 - P (X = 0)
= 1 - [e^-(np) ][(np)^0] / [0!]
= 1 - e^-(np)
= 1 - 1.0000007665
= well.. u know the result....