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Bernoulli trials and limit theorems

Hence, we have that np − q ≤ k 0 ≤ np + p is valid in general case. 2

Example 4.4. What are most probable number of getting a ’six’ in 5 tosses of
dice? ,

Solution. We have n = 5, p = 1/6, q = 5/6, np = 5/6, np − q = 0, np + p = 1. Hence
0 ≤ k 0 ≤ 1. Thus there are two most probable values of getting a ’six’: k 0 = 0 and k 0 = 1.

Limit Poisson Theorem

If we have experiments by Bernoulli trials and number n and k are large, then evaluating
probabilities P n (k) by Bernoulli formula is very difficult. It is usefully to apply asymptotic
formulas for evaluating of this probabilities. These formulas are implied de Moivre-Laplace

theorem and Poisson limit theorem. These theorems describes an behaviour of P n (k) and P n
(k 1 ≤ k ≤ k 2 ) for some conditions, when n → ∞.

Theorem 4.2. (Limit Poisson Theorem). If p is an probability of success in every of n
independent experiments by Bernoulli trials and if for n → ∞ p → 0 such that np → λ
(0 < λ < ∞), then
λ k
lim P n (k) = e −λ
n→∞ k!
for all k = 0, 1, 2, . . . , where P n (k) is an probability of k successes in n experiments. ⋆

PROOF. We put np = λ n (i. e. λ n → λ for n → ∞) and write

k
( ) ( ) n−k
n! n(n − 1) · · · (n − k + 1) λ n λ n
k n−k
k k n−k
P n (k) = C p q = p q = · 1 − =
n
(n − k)!k! k! n n
n
λ k ( λ n ) ( 1 ) ( 2 ) ( k − 1 ) ( λ n ) −k
= n 1 − 1 − 1 − · · · 1 − 1 − −−−−→
n → ∞
k! n n n n n
[ ] −λ
( ) n k ( ( )) − n k
k
λ λ λ λ λ
−−−−→ λ n −λ
n → ∞ · lim 1 − = · lim 1 + − = e .
k! n→∞ n k! n→∞ n k!
Theorem is proved. 2
Thus, for large n (n > 100) and small p (np < 30) we can use this approximate formulas:
λ k
P n (k) ≈ e −λ ; (4.5)
k!

k 2 k 2 k
∑ ∑ λ
P n (k 1 ≤ k ≤ k 2 ) = P n (k) ≈ e −λ , λ = np, (4.6)
k!
k=k 1 k=k 1
The relationships (4.5) and (4.6) are called asymptotic Poisson formulas. A second
relationship gives an approximate estimation of probability that a number of successes in n
experiments lies between k 1 and k 2 .
We can use a following inequality for estimation of accuracy (4.5) and (4.6)

k
∑ ∑ λ
−λ 2
P n (k) − e < np ,

k!
k∈M k∈M
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