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midterm_old_sol - NTHU MATH 2810 Midterm Examination...

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NTHU MATH 2810 Midterm Examination Solution Oct 30, 2007 1. (a) ( 3pts ) Let n i be the number of candies the i th child got, where i = 1 , 2 , 3 , 4 , then it must be satisfied that n i 1 , i = 1 , 2 , 3 , 4 , and n 1 + n 2 + n 3 + n 4 = 10 . (1) The number of integer solutions for Eqn. (1) is 10 - 1 4 - 1 · = 84 . (b) ( 3pts ) For each of the 10 different books, there are four distinct (and independent) choices of child it can be given to. So, the answer is 4 × 4 × · · · × 4 = 4 10 . 2. ( 8pts ) The three equations can be represented in terms of p 1 , p 2 , p 3 , and p 4 as follows: P ( A B ) = 3 P ( B ) p 1 + p 2 + p 3 = 3( p 1 + p 3 ) , (2) P ( A B ) = 0 . 4 P ( A B c ) p 1 = 0 . 4 p 2 , (3) P (( A B ) c ) = 0 . 1 p 4 = 0 . 1 . (4) Furthermore, because ( A B ) ( A B c ) ( A c B ) ( A c B c ) = Ω (the whole sample space) , p 1 + p 2 + p 3 + p 4 = 1 . (5) By solving Eqns. (2), (3), (4), and (5), we get p 1 = 0 . 24 , p 2 = 0 . 6 , p 3 = 0 . 06 , and p 4 = 0 . 1 . Therefore, P ( A ) = p 1 + p 2 = 0 . 84 . 3. (a) ( 4pts ) Let E be the event of picking a black ball the first draw. Then, P ( E | U 1 ) = 3 5 and P ( E | U 2 ) = 2 5 . So, by the law of total probability, P ( E ) = P ( E | U 1 ) · P ( U 1 ) + P ( E | U 2 ) · P ( U 2 ) = 3 5 · 1 2 + 2 5 · 1 2 = 1 2 .
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