Binomial Distribution: Average & Probability of ≥1 Success

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The average of a binomial distribution with success probability k is given by the formula Nk. The probability of achieving at least one success in N trials can be approximated as P(≥1) = Nk when Nk is sufficiently small. This approximation does not hold when Nk exceeds 1 or for small values of N. The probability can also be expressed as P(at least one success) = 1 - (1-k)N, which highlights the relationship between successes and failures. The discussion emphasizes the need for Nk to be small for the approximation to be valid, while also noting potential errors in notation.
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The average if the binomial distribution with probability k for succes is simply:

<> = Nk

So this means that if <> = 1 the distribution function must be peaked around 1. In general when is it a good approximation (i.e. when is the function peaked sufficiently narrow) to say that the probability in N tries to have one or more succeses is simply:

P(≥1) = Nk

this obviously does not hold for Nk>1 but on the other hand I don't expect it to hold for small N. So my guess is when Nk is sufficiently small. Is that correct?
 
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P(at least one success) = 1 - P(all failure) = 1 - (1-k)N.
You should be able to analyze it.
 
Right so using the binomail theorem you find:

P(at least one success) = Nk - K(N,2)k2 + K(N,3)k3 - K(N,4)k4 + ...

So the question is when the first term dominates. I am guessing for sufficiently small k?
 
You need Nk small, not just k. In your expression you seem to have extraneous K (capital k) from the second term on.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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