P(E|F) (1)
Also, we know that
P(F|E) = P(F
P(E)E) ,P(E) 0
∩
≠
or
P(F|E) = P(E
P(E)F)
∩
 (since E ∩ F = F ∩ E)
Thus,
P(E ∩ F) = P(E) P(F|E) (2)
Combining (1) and (2), we find that
P(E ∩ F) = P(E) P(F|E)
= P(F) P(E|F) provided P(E) ≠ 0 and P(F) ≠ 0