Page 105 - MATINF Nr. 13-14

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Rezolvarea problemelor date la concursul de admitere la Ecole Polytechnique, filiera MPI, sesiunea 2024 105

Not˘am

p = P (Y n = −1) , q = P (Y n = 1) = 1 − p.

Å ã
−1 1
Repartit , ia variabilei aleatoare Y n este Y n : .
p q
7. b. Dac˘a t , inem cont c˘a
n k
X (−1)
lim = e −1
n→∞ k!
k=0

s , i
n−1
(−1) (n − 1)
lim = 0
n→∞ n!

obt , inem c˘a

 
n−1 Å ã
1  1 (−1) (n − 1)  1 1 1
lim P (Y n = −1) = lim  1 − ·  = 1 − · 0 =
n
n→∞ 2 n→∞  P (−1) k n!  2 e −1 2
k!
k=0
s , i

1 1
lim P (Y n = 1) = 1 − lim P (Y n = −1) = 1 − = .
n→∞ n→∞ 2 2

8. a. Am ar˘atat la punctul 6 c˘a num˘arul permut˘arilor cu k puncte fixe (ν (σ) = k) este
k
C · D n−k . Prin urmare, folosind relat , ia de la punctul 6,
n
n−k l n−k l
X (−1) n! X (−1)
k
card {Z n = k} = card {σ ∈ S n : ν (σ) = k} = C · (n − k)! · = · .
n
l! k! l!
l=0 l=0
De aici

n−k l
card {Z n = k} 1 X (−1)
P (Z n = k) = = · .
cardS n k! l!
l=0
Ç å
k
Not˘am p k = P (Z n = k) s , i repartit , ia variabilei aleatoare Z n este Z n : .
p k
k=0,n
8. b. Avem

n−k l ∞ l
1 X (−1) 1 X (−1) 1
−1
lim P (Z n = k) = lim · = · = · e .
n→∞ n→∞ k! l! k! l! k!
l=0 l=0