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math 105 Prelim2 +key 2004

# math 105 Prelim2 +key 2004 - Math 105 Fall 2004 Solutions...

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Math 105, Fall 2004 – Solutions to Prelim 2 1. (a) Henri has to take 3 bottles from 5+7=12, and the order in which they are taken doesn’t matter. Therefore, there are 12 3 = 12! 9! 3! = 12 · 11 · 10 · 9! 9! 3! = 12 · 11 · 10 3! = 1320 6 = 220 possible choices. (b) Henri has to choose 1 bottle from the 5 bottles of red and 2 bottles from the 7 bottles of white. By the multiplication principle, there are 5 1 · 7 2 = 5 · 21 = 105 ways of doing this. 2. Note that S = { 0 , 1 , 2 , . . . , 98 , 99 } is the sample space for this problem. (a) E = { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 } . F = { 9 , 18 , 27 , 36 , 45 , 54 , 63 , 72 , 81 , 90 } . Assuming equally likely outcomes, we have: P ( E ) = n ( E ) n ( S ) = 20 100 = 1 5 and P ( F ) = n ( F ) n ( S ) = 10 100 = 1 10 . (b) Assuming equally likely outcomes, P ( E F ) = n ( E F ) n ( s ) = n ( { 9 , 18 } ) 100 = 2 100 = 1 50 . So by definition of conditional probability, we have: P ( E | F ) = P ( E F ) P ( F ) = 1 / 50 1 / 10 = 1 5 . Hence P ( E ) = 1 5 = P ( E | F ), and so E and F are independent events.

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(c) G = { 5 , 14 , 23 , 32 , 41 , 50 } . So assuming equally likely outcomes, we have P ( G ) = n ( G ) n ( S ) = 6 100 = 3 50 and P ( E G ) = n ( E G ) n ( S ) = n ( { 5 , 14 } ) 100 = 2 100 = 1 50 . Hence P ( G | E ) = P ( E G ) P ( E ) = 1 / 50 1 / 5 = 1 10 . So P ( G | E ) = 1 10 = 3 50 = P ( G ) and therefore E and G are not independent.
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