Question 1162428: P(E)=0.35, P(F)=0.35, P(E ∩ F)=0.15 and P((E ∪ F)c)=0.45. Find P(E ∩ FC|F)
Answer by ikleyn(52802)   (Show Source):
You can put this solution on YOUR website! .
OK.
Now i got everything in your post and in your designations (which you failed to disclose in the post).
(1) The symbol "c" means "the complement" in each appearance.
(2) The equality P((E ∪ F)c)=0.45 is excessive and unnecessary,
since it follows from the preceding data:
P(E)=0.35, P(F)=0.35, P(E ∩ F)=0.15 implies P(E U F) = P(E) + P(F) - P(E ∩ F) = 0.35 + 0.35 - 0.15 = 0.55,
which is exactly the same as P((E ∪ F)c)=0.45.
(3) Finally, P(E ∩ FC|F) = by the definition = P( (E ∩ FC) ∩ F) / P(F) = 0,
since the set (E ∩ FC) ∩ F is empty.
ANSWER. P(E ∩ FC|F) = 0.
Solved, deciphered, answered and completed.
The reader may consider it as a joke problem . . .
The only thing that remained unclear to me is if the author of this post did understand that it was a joke problem . . .
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