Question:
Prove that if \(E\) and \(F\) are independent events, then \(E\) and \(F'\) will also be independent.
Prove that if \(E\) and \(F\) are independent events, then \(E\) and \(F'\) will also be independent.
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Write P(E∩F') = P(E) − P(E∩F) and substitute P(E∩F) = P(E)P(F).
Updated On: Sep 23, 2026
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Solution and Explanation
Step 1: 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 2: Isolating P(E∩F'):
\(P(E\cap F')=P(E)-P(E\cap F)\).
Step 3: Using independence of E, F:
Since \(E,F\) are independent, \(P(E\cap F)=P(E)P(F)\), so \(P(E\cap F')=P(E)-P(E)P(F)=P(E)[1-P(F)]\).
Final Answer:
Since \(1-P(F)=P(F')\), we get \(P(E\cap F')=P(E)P(F')\), which is exactly the definition of \(E,F'\) being independent.\[ \boxed{E \text{ and } F' \text{ are independent}} \]
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 2: Isolating P(E∩F'):
\(P(E\cap F')=P(E)-P(E\cap F)\).
Step 3: Using independence of E, F:
Since \(E,F\) are independent, \(P(E\cap F)=P(E)P(F)\), so \(P(E\cap F')=P(E)-P(E)P(F)=P(E)[1-P(F)]\).
Final Answer:
Since \(1-P(F)=P(F')\), we get \(P(E\cap F')=P(E)P(F')\), which is exactly the definition of \(E,F'\) being independent.\[ \boxed{E \text{ and } F' \text{ are independent}} \]
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