sec_2_3_sol

sec_2_3_sol - APPM 4/5570 Solutions to Problems from...

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Unformatted text preview: APPM 4/5570 Solutions to Problems from Section 2.3 30. a. P 3 , 8 = 8! (8- 3)! = 8! 5! = (8)(7)(6) = 336 b. C 6 , 30 = 30 6 ! = 30! 6! (30- 6)! = 593 , 775 c. You must choose 2 of the 8 bottles of zinfandel, 2 of the 10 bottles of merlot, and 2 of the 12 bottles of cabernet. So, the answer is: 8 2 ! 10 2 ! 12 2 ! = 28 · 45 · 66 = 83 , 160 d. In part (c) we detrmined the total number of ways to select six bottles while specifically getting 2 of each kind. In part (b) we determined the total number of ways to select 6 bottles with no further restrictions. The answer is the answer in (c) over the answer in (b): 8 2 ! 10 2 ! 12 2 ! 30 6 ! = 83 , 160 593 , 775 ≈ . 14 e. We can have all zinfandel or all merlot or all cabernet. Since these are disjoint events, we get the desired probability by adding together the three corresponding probabilities. I have put the “zero factors” in for complet- ness, but since “10 choose 0”, for example, is equal to 1, they do not need to be there....
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This note was uploaded on 01/23/2011 for the course APPM 5440 at Colorado.

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sec_2_3_sol - APPM 4/5570 Solutions to Problems from...

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