20-hw6

# 20-hw6 - Stat 20: solutions to homework 6 Michael Lugo...

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Unformatted text preview: Stat 20: solutions to homework 6 Michael Lugo October 12, 2010 1 Chapter 15, # 5 The second list is longer than the first. The first list contains 8 2 ! = 8! 2!6! = 8 × 7 2 × 1 = 28 people and the second list containts 8 5 ! = 8! 3!5! = 8 × 7 × 6 3 × 2 × 1 = 56 people. 2 Chapter 15, # 8 The two events “2 heads among the first 5 tosses” and “4 heads among the last 5 tosses” are independent. The first one happens with probability 5 2 (1 / 2) 5 = 10 / 32 and the second with probability 5 4 (1 / 2) 5 = 5 / 32. The probability we want is the product, (10 / 32)(5 / 32) = 50 / 1024. 3 Chapter 15, # 11 (a) This is the sum of the probabiliy that k out of 22 pairs have the smoker die first, for k = 17 , 18 ,..., 22. This is 22 17 + 22 18 + 22 19 + 22 20 + 22 21 + 22 22 2 22 = 26334 + 7315 + 1540 + 231 + 22 + 1 4194304 1 which is 35443 / 4194304, or about 0 . 00845. (b) This is (1 / 2) 9 = 1 / 512. (c) This is (1 / 2) 2 = 1 / 4....
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## This note was uploaded on 07/21/2011 for the course STAT 20 taught by Professor Staff during the Spring '08 term at Berkeley.

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20-hw6 - Stat 20: solutions to homework 6 Michael Lugo...

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