1 Hicham et SAni · · · 1.1 Dé nition

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P(A)̸= 0 B A
pA(B)
pA(B) = P(A∩B)
P(A)
B A
pA(B) = P(A∩B)
P(A)=A∩B A
pC(M)
pM(C)
pM(C) = 12
34
A B
A A B
P(A∩B) = P(A)×pA(B)
A B Ω
A¯ AΩ
A¯ A
B
P(B) = P(A∩B) + P(¯ A∩B)
ΩA1, A2, . . . , Am
E E AiAi
A1∪A2∪ · · · ∪ Am=E
Ai∩Aj=∅i, j m
P(B) = P(B∩A1) + P(B∩A2) + · · · +P(B∩Am)

A
B
C
P(A)P(B)P(A∩B)P(C)P(B∩C)
P(A) = 4
32 =1 8P(B) = 8 32 =1 4P(C) = 4 32 =1 8
B∩C
P(B∩C) = 1
32 =1 4×1 8=P(B)×P(C)
B C
P(A∩B) = P(A)×P(B)
P(A∩B) = P(A)×P(B)pB(A) = P(A)pA(B) = P(B)
A B
A B A
A B
P(A∩B) = ∅
A
P(A) = p A P (¯ A) = q¯ A p
q p +q= 1
A
P(A) = 1/6P(¯ A) = 5/6
A¯ A
n
n p
p
n

n
k
k4−3 = 1 n−k
P(12) = (1/6)3×(5/6)1=5
1296 = 0,00385 . . . P (k) = (1/6)k×(5/6)n−k
n n ≥1p∈[0,1]
n p
E={0,1,2,3, . . . , n}
E n p
B(n, p)
k n −k n
P(k) = pk×qn−k
{0,1,2,3, . . . , n}
P(k) = Ck
n×pk×qn−k