SOLUTION: Suppose P(A)=0.3 and P(B)=0.4 and P(A∩B)=0.12. Find: a. P(B') b. P((A∩B)') c. P((A∪B)') ('=complement)

Algebra ->  Probability-and-statistics -> SOLUTION: Suppose P(A)=0.3 and P(B)=0.4 and P(A∩B)=0.12. Find: a. P(B') b. P((A∩B)') c. P((A∪B)') ('=complement)      Log On


   

Question 1116432: Suppose P(A)=0.3 and P(B)=0.4 and P(A∩B)=0.12. Find:
a. P(B')
b. P((A∩B)')
c. P((A∪B)')
('=complement)
Found 2 solutions by stanbon, ikleyn:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose P(A)=0.3 and P(B)=0.4 and P(A∩B)=0.12. Find:
a. P(B') = 0.6
b. P((A∩B)') = 0.88
c. P((A∪B)')
P(A or B) = 0.18+0.12+0.28 = 0.58
So, P(A or B)' = 0.42
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Cheers,
Stan H.
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Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
Suppose P(A)=0.3 and P(B)=0.4 and P(A∩B)=0.12. Find: 

a. P(B')                P(B') = 1 - P(B) = 1 - 0.4 = 0.6.


b. P((A∩B)')            P((A∩B)') = 1 - P(A∩B) = 1 - 0.12 = 0.88.


c. P((A∪B)')            First  P(A∪B) = P(A) + P(B) - P((A∩B) = 0.3 + 0.4 - 0.12 = 0.58

                        Then   P((A∪B)') = 1 - P(A∪B) = 1 - 0.58 = 0.42.


('=complement)