Loi binomiale, cours, première ES Table des - MathsFG

p q
q= 1 −p n
n p
p= 0,05
n= 3 p= 0,05
P(SS ¯ S) = 0,052×0,95 ≈0,002

p p ∈[0; 1] X X
p
•X
•P(X= 1) = p P (X= 0) = 1 −p
X p E(X) = p
E(X)=0×P(X= 0) + 1 ×P(X= 1) = 0 ×(1 −p)+1×p=p
n k k ≤nn
kk n
k n −k n
4
4= 1
4
3= 4
SSSE SSES SESS ESSS

p∈[0; 1] n n ≥1
X
X n
p X ∼ B(n;p)
X n p
k k ∈ {0; 1; 2; ...;n}P(X=k) = n
kpk(1 −p)n−k
n
kk n
pk(1 −p)n−k
n= 3 p= 0,05 X
n= 3 p= 0,05
P(X= 2) = P(SS ¯ S∩S¯ SS ∩SS ¯ S)
P(X= 2) = P(SS ¯ S) + P(S¯ SS) + P(SS ¯ S)
P(X= 2) = 0,052×0,95 + 0,95 ×0,05 ×0,95 + 0,95 ×0,052
P(X= 2) = 3 ×0,052×0,95 = 3
20,052(1 −0,05)3= 0,007
P(X=k)P(X≤k)
X n p k n P (X=k)
P(X≤k)
•
P(X=k)P(X≤k)
•
P(X=k)P(X≤k)
•P(X < 3) = P(X≤2)
•P(X > k) 1 −P(X≤k)

n p
E(X) = np
n= 100
n= 100 p= 0,05 E(X) = np = 100 ×0,05 = 5