SOLUTION: Let P(E) = 0.45, P(F) = 0.55, and P(F ∩ E) = 0.25. Draw a Venn diagram and find the conditional probabilities. (a) P(E | F^C ) (b) P(F | E^C )

Algebra ->  Probability-and-statistics -> SOLUTION: Let P(E) = 0.45, P(F) = 0.55, and P(F ∩ E) = 0.25. Draw a Venn diagram and find the conditional probabilities. (a) P(E | F^C ) (b) P(F | E^C )      Log On


   

Question 1203775: Let P(E) = 0.45, P(F) = 0.55, and P(F ∩ E) = 0.25. Draw a Venn diagram and find the conditional probabilities.
(a) P(E | F^C )

(b) P(F | E^C )
Answer by ikleyn(52781) About Me  (Show Source):
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Let P(E) = 0.45, P(F) = 0.55, and P(F ∩ E) = 0.25. Draw a Venn diagram and find the conditional probabilities.
(a) P(E | F^C )
(b) P(F | E^C )
~~~~~~~~~~~~~~~~~~~~~~~~~ 

(a)  Conditional probability  P(E | F^C )  is

         P(E | F^C ) = P%28E_and_F%5Ec%29%2FP%28F%5Ec%29


     The intersection  E_and_F^c  consists of those elements of E that do not belong to F.

     Hence, the set  E_and_F^c  is  the same as  E \ (E ∩ F).

     Therefore,  P(E_and_F^c) = P(E) - P(E ∩ F) = 0.45 - 0.25 = 0.2.

     From the other side,  P(F^c) = 1-P(F^c) = 1 - P(F) = 1 - 0.55 = 0.45.

     Thus  P(E | F^C ) = P%28E_and_F%5Ec%29%2FP%28F%5Ec%29 = 0.2%2F0.45 = 20%2F45 = 4%2F9.    ANSWER


The solution for (b) follows the same logic. 

(b)  The answer for (b) is  %280.55-0.25%29%2F%281-0.45%29 = 0.3%2F0.55 = 30%2F55 = 6%2F11.

Solved.