{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Sol_ma3160a10809B - ma3160a1 ma4535a1 ma5160a1 ma5610a1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}