P(B∩A) | ||
Wiem, że | =p | |
P(A) |
P(A∩B) | |
= q | |
P(B) |
(1−r)*q | ||
P(A)= | ||
q−pq+p |
(1−r)*p | ||
P(B)= | ||
q−pq+p |
p*P(A) | ||
P(A∩B) =p*P(A) i P(A∩B)=q*P(B) ⇒ P(B)= | ||
q |
p*P(A) | ||
P(A)+ | −p*P(A)= 1−r /*q | |
q |
(1−r)*q | ||
P(A)(q+p−pq)=(1−r)*q ⇒ P(A)= | ||
p−pq+q |
p | ||
P(B)= | *P(A)=................ | |
q |