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HW 3 Solutions

# HW 3 Solutions -...

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1 http://www.stat.ucla.edu/~dinov/courses_students.dir/11/Winter/STAT13.1.dir STAT 13, section 1, Winter 2011, UCLA Statistics HW 3; Problem Solution HW 3.1 First, define the notations: M ={A male child born} (could have been F={A female child born}. No intention of gender discrimination) D={Have the disease (Affected)} Then, from the problem: P(M)=0.51 P(D|M)=0.5 ; P(D|M c )=0 From the given information, we can infer that P(D M) = P(D|M)P(M) = 0.255 P(D M c ) = P(D|M c )P(M c )=0 (a) Since we are told that 2 children are a male and a female, P(a male affected a female affected) = P(M ∩D) ∙ P(M c ∩D) = 0.255 ∙ 0 = 0 (b) In this problem, we want to get P(DD c ). Since we are not given the gender of the 2 children, we have to consider 3 cases; male and male, male and female, and female and female. First, using binomial distribution with n=2 and p=0.51, we have P(male and male)=P(MM)= 2601 . 0 ) 51 . 0 1 ( 51 . 0 2 2 0 2 = P(MM c )= 4998 . 0 ) 51 . 0 1 ( 51 . 0 1 2 1 1 = P(M c M c )= 2401 . 0 ) 51 . 0 1 ( 51 . 0 0

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HW 3 Solutions -...

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