Solution_Assign2_Fall_2010

Solution_Assign2_Fall_2010 - Solutions for Exercise 3 1....

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Solutions for Exercise 3 1. With n 1 =6 sightseeing tours each available on n 2 =3 different days, the multiplication rule gives n 1 n 2 =(6)(3)=18 ways for a person to arrange a tour. 2. Since the die can land in n 1 =6 ways and a letter can be selected in n 2 =6 ways, the multiplication rule gives n 1 n 2 =(6)(26)=156 points in S. 3. P 8 5 = 8! 3! =6720. 4. P 40 3 = 40! 37! =59280. 5. There are 7!‘=5040 arrangements. 6. There are ( 12 7 , 3 , 2 )=7920 ways. 7. ( 9 1 , 4 , 4 ) + ( 9 2 , 4 , 3 ) + ( 9 1 , 3 , 5 ) + ( 9 2 , 3 , 4 ) + ( 9 2 , 2 , 5 )=4410. 8. P ( S | A ) = 10 / 18 = 5 / 9 9. Consider the events : M: a person is a male, S: a person has a secondary education, C: a person has a college degree. a) P ( M | S ) = 28 / 78 = 14 / 39 b) P ( C 0 | M 0 ) = 95 / 112. 10. a) P ( M T P T H | P )=10/68=5/34. b) P ( H T M | P 0 ) = 12 / 32 = 3 / 8 . 11. a) 0.018 b) 0.022+0.002+0.160+0.046+0.084=0.614 c) 0.102/0.614=0.166 d) (0.102+0.046)/(0.175+0.134)=0.479. 12. Consider the events:
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This note was uploaded on 10/20/2011 for the course MATH 144 taught by Professor Yu during the Spring '11 term at HKU.

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Solution_Assign2_Fall_2010 - Solutions for Exercise 3 1....

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