Step 1: Key Approach:
We must show \(P(E\cap F')=P(E)\cdot P(F')\), using the fact that \(P(E\cap F)=P(E)P(F)\) since \(E,F\) are independent.
Step 2: Splitting E using F and F':
Since \(F\) and \(F'\) partition the sample space, \(E=(E\cap F)\cup(E\cap F')\), and these two pieces are disjoint. So \(P(E)=P(E\cap F)+P(E\cap F')\).
Step 3: Solving for P(E∩F'):
\(P(E\cap F')=P(E)-P(E\cap F)=P(E)-P(E)P(F)=P(E)\big[1-P(F)\big]\).
Step 4: Recognizing 1-P(F) as P(F'):
By the complement rule, \(1-P(F)=P(F')\). So \(P(E\cap F')=P(E)\cdot P(F')\).
Final Answer:
This is exactly the condition for independence of \(E\) and \(F'\).\[ \boxed{P(E\cap F')=P(E)P(F')\ \Rightarrow\ E,F'\text{ independent}} \]