# DSE 2011 Queries Classic List Threaded 10 messages Open this post in threaded view
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## DSE 2011 Queries

 Question 9. Could someone explain how we obtain this probability: (1-p1)(1-p2)(1-p3) Question 25. How is it discontinous at all points for x =/ 0? Q27-30: Could someone explain these logic questions? Also question 37. Any help regarding these questions would be greatly appreciated :) http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/2011-option-a.pdf
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## Re: DSE 2011 Queries

 Could anyone help? I have the same doubts as Ayushya. Also I need help with Q4) and Q5) of DSE 2011. Any help is welcome people :)
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## Re: DSE 2011 Queries

 In reply to this post by Ayushya Kaul Hi Ayushya.. :) 9) P(None of them occurs) = P(ȦḂĊ) = P(Ȧ)*P(Ḃ)*P(Ċ) [As A,B,C are independent and its easy to show that Ȧ,Ḃ,Ċ will also be independent. ] Therefore, P(None of them occurs) = (1-p1)(1-p2)(1-p3) Note: Ȧ is complement of A, Ḃ is complement of B and Ċ is complement of C. 25) Use the definition of continuity. :)
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## Re: DSE 2011 Queries

 In reply to this post by Ayushya Kaul 27) C1 ∩ C2 = (C1' ∪ C2')' Note: C1' denotes complement of C1. Now, C1' is a tribe and C2' is also a tribe. And union is two tribes is a tribe therefore, (C1' ∪ C2') is a tribe. Hence, complement of (C1' ∪ C2') = (C1' ∪ C2')' will be a club.Hence, option(a) 28) C1 ∪ C2 = (C1' ∩ C2')' Now, C1' and C2' are tribes. So, their intersection will also be a tribe. And therefore, the complement of (C1' ∩ C2') = (C1 ∩ C2)' will be a club. Hence, option(c) 29) Society and Empty set are both Club and Tribe.Hence, option(b) 30)Given a tribe T1 and a club C2: (T1 ∩ C2') is a club since C2' is  a tribe and the intersection of two tribes is a tribe. Hence, option(b) :)
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## Re: DSE 2011 Queries

 In reply to this post by Ayushya Kaul its a problem of multicollinearity . we go by options . a) can't be as x1x2 will be correlated with x1 and x2. b) ans as there won't be any correlation b/w d exogenous variables c)correlation b/w x1x2 and x1 and x2 . there is perfect collinearity and hence d parameters can't be estimated .
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## Re: DSE 2011 Queries

 4) Given: a1.x1 + a2.x2 + a3.x3 +....an.xn = b ⇒ a1.x1 + a2.x2 + a3.x3 +....an.xn - b = 0 [Note: RHS = 0 . Here, "0" is a vector , not a real number]. So, its like taking a linear combination of vectors {a1,a2...an,b} and  as "x" is a non-zero vector then there is some x(i) which is not equal to zero. Therefore, the set of vectors {a1,a2...an,b} is linerly dependent. :)
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## Re: DSE 2011 Queries

 In reply to this post by Darth Vader Hi Darth, For question 5, as m>n hence number of equations is more than number of variables. Therefore, if we can find a set of n equations amongst these m equations such that the A matrix is full rank. Then its inverse exists and hence there will be a unique solution.
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## Re: DSE 2011 Queries

 In reply to this post by duck Hey Duck :) Thanks a lot! That was really helpful, cleared up a lot of things for me. Couldn't thank you less. Once again- Cheers :D