| 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 |
na rozluźnienie szarych komórek