Idée de corrigé

X
Y Z
D B
p(X) = 15
100 = 0,15
D X
p(D|X) = 5
100 = 0,05
D∩X
p(D∩X) = p(D|X)×p(X)
= 0,0075
p(D∩Y)p(D∩Z)
p(D∩Y) = p(D|Y)×p(Y)
= 0,03 ×0,55
= 0,0165
p(D∩Z) = p(D|Z)×p(Z)
= 0,02 ×0,30
= 0,0060
X Y Z
p(D) = p(D∩X) + p(D∩Y) + p(D∩Z)
= 0,030
p= 0.030
p(D∩X) = 0,0075 p(D)×p(X) = 0,0045
X D
p(Z|D)×p(D) = p(Z∩D)
p(Z∩D) = 0,0060 p(D) = 0,030
p(Z|D) = p(Z∩D)
p(D)
= 0,20

p(X|D)p(Y|D)p(Z|D)
p(X|D) = p(X∩D)
p(D)
= 0,25
p(Y|D) = p(Y∩D)
p(D)
= 0,55
D
S D R
S
p(S= 2) = p(X|D) = 0,25
p(S= 3) = p(Y|D) = 0,55
p(S= 5) = p(Z|D) = 0,20
S
S= 2 ×p(S= 2) + 3 ×p(S= 3) + 5 ×p(S= 5)
= 3,15
R
R
R
p(R= 0) = p(D) = 0,970
p(R= 2) = p(X∩D)=0,0075
p(R= 3) = p(Y∩D) = 0,0165
p(R= 5) = p(Z∩D) = 0,0060
R
R= 2 ×p(R= 2) + 3 ×p(R= 3) + 5 ×p(R= 5)
= 0,0945
S= 3,15
p(D) = 0,030
R= 0,030 ×3,15 = 0,0945

p(X)4
p4=p(X)4+p(Y)4+p(Z)4
= 0,1001125
p= 0,030
q= 1 −p
M N = 4 −M
M N
p(M=k) = Ck
4pkqn−kk∈[0..4]
p(M=k)=0 k /∈[0..4]
p(N=k) = Ck
4qkpn−kk∈[0..4]
p(N=k) = 0 k /∈[0..4]
ξ
ξ=p(N= 3) + p(N= 4)
=C3
4q3p+C4 4q4
= 4 ×(1 −0,03)3×0,03 + (1 −0,03)4
ξ
ξ=p(M= 0) + p(M= 1)
=C0
4q4+C1 4p q3
= (1 −0,03)4+ 4 ×0,03 ×(1 −0,03)3
ξ= 0,99481357