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Sol_ma3160a10809B

# Sol_ma3160a10809B - ma3160a1 ma4535a1 ma5160a1 ma5610a1...

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ma3160a1 ma4535a1 ma5160a1 ma5610a1 Probability 1. Consider two events A and B such that ( 29 0.4 P A = and ( 29 0.7 P B = . Determine the maximum and minimum possible values of ( 29 P A B and the conditions under which each of the value is attained. Solution: ) ( ) ( ) ( ) ( B A P B P A P B A P - + = . 7 . 0 )} ( ), ( { max ) ( and ) ( ), ( ) ( , , = B P A P B A P B P A P B A P B A B A B In addition, ). ( 1 B A P Thus . 7 . 0 ) ( 1 B A P So if , 1 ) ( = B A P ) ( B A P will attain the minimum and 1 . 0 1 7 . 0 4 . 0 ) ( = - + = B A P If , 7 . 0 ) ( = B A P ) ( B A P will attain the maximum and 4 . 0 7 . 0 7 . 0 4 . 0 ) ( = - + = B A P 2. Let , , A B C be events, prove that ( 29 29 ( 29 ( ( 29 1 c c c c c c P A B C P A B C P B C P C = - Proof: ( 29 ( 29 ( 29 29 ( 29 ( ( 29 1 1 c c c c c c c c c c c c c P A B C P A B C P A B C P A B C P B C P C = = - = - 3. Let ( 29 P , , Γ be a probability space. Let , A B ∈Γ and ( 29 ( 29 1 P A P B = = . Find (a) ( 29 c P A B , (b) ( 29 P A B , ( 29 P A B , (c) ( 29 c P A B . (d) If , , A B C ∈Γ are independent, show that , A B C are independent Solution: (a) ( 29 ( 29 ( 29 1 1 0 c P B P B P B = = - = , ( 29 ( 29 ( 29 0 0 0 c c c c c A B B P A B P B P A B = = . (b) ( 29 ( 29 ( 29 1 1 1 A A B P A P A B P A B = = . And ( 29 ( 29 ( 29 ( 29 1 1 1 1 P A B P A P B P A B = + - = + - = . (c) ( 29 ( 29 ( 29 1 1 1 c c c A A B P A P A B P A B = = (d) ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 , , are independent , are independent A B C B C P A B C P A B C P A P B P C P A P B C = = = 4. Suppose events A and 1 B , events A and 2 B are independent , the events 1 B and 2 B being disjoint. Prove that the events A and 1 2 B B are independent. Proof: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 2 1 2 1 2 1 2 1 2 1 2 P A B B P A B A B P A B P A B P A P B P A P B P A P B P B P A P B B = = + = + = + = 5. Let events A and B be independent and A B . Prove that either ( 29 0 P A = or ( 29 1 P B = . Proof: A B A A B = . Thus ( 29 ( 29 ( 29 ( 29 P A P A B P A P B = = , events A and B are independent. So ( 29 ( 29 1 0 P A P B - = . It follows that ( 29 ( 29 0 or 1 0 P A P B = - = , that is, ( 29 0 P A = or ( 29 1 P B = . 6. (a) If A is independent of itself, show that ( 29 P A is 0 or 1. (b) If ( 29 P A is 0 or 1, show that A is independent of all events B . 1

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Proof: (a) ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 0 1 0 0 or 1 0 0 or 1 P A P A A P A P A P A P A P A P A P A P A P A P A P A = = - = - = = - = = = (b) Suppose that ( 29 0 P A = . ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 0 0 0 0 A B A P A B P A P A B P A B P A P B = = = = . Suppose that ( 29 1 P A = . ( 29 ( 29 ( 29 1 1 1 A A B P A P A B P A B = = . Since ( 29 ( 29 ( 29 ( 29 P A B P A P B P A B = + - and ( 29 ( 29 1 P A P A B = = , we have ( 29 ( 29 ( 29 ( 29 P A B P B P A P B = = 7. Among a group of 200 students, 137 students are enrolled in a mathematics class, 50 students in a history class and 124 students in a music class. Furthermore, the number of students enrolled in both the mathematics and history classes is 33; the number of students enrolled in both the history and music classes is 29; the number of students enrolled in both the mathematics and music classes is 92.
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