Romanov Stole My Pension Posted October 22, 2008 Share Posted October 22, 2008 Long shot but hopefully some statistics students read this. Question is... Let A, B and C be events. Show that if P(A|C)>P(B|C) and P(A|C[complement])>P(B|C[complement]), then P(A)>P(. Link to comment Share on other sites More sharing options...
Seats Posted October 22, 2008 Share Posted October 22, 2008 37. You're welcome Link to comment Share on other sites More sharing options...
I P Knightley Posted October 22, 2008 Share Posted October 22, 2008 It almost goes without saying Link to comment Share on other sites More sharing options...
Eggo Posted October 22, 2008 Share Posted October 22, 2008 Long shot but hopefully some statistics students read this.Question is... Let A, B and C be events. Show that if P(A|C)>P(B|C) and P(A|C[complement])>P(B|C[complement]), then P(A)>P(. More P's required to be honest,or the answer will B a B..U C. Link to comment Share on other sites More sharing options...
Jamboelite Posted October 22, 2008 Share Posted October 22, 2008 Long shot but hopefully some statistics students read this.Question is... Let A, B and C be events. Show that if P(A|C)>P(B|C) and P(A|C[complement])>P(B|C[complement]), then P(A)>P(. Come on even a first year stats student can see the answer. Dont bother me unless you have a more complex problem. Link to comment Share on other sites More sharing options...
Romanov Stole My Pension Posted October 22, 2008 Author Share Posted October 22, 2008 I know it's obvious to see that it makes sense. But i have no idea how to write down and prove that it is true. Link to comment Share on other sites More sharing options...
Commander Harris Posted October 22, 2008 Share Posted October 22, 2008 p(A|C) = p(A[intersect]C)/p© p(B|C) = p(B[intersect]C)/p© similarly for c compliment Express inequalities in that form. Cancel the fractions ( muliply by p© or p(C[compliment) ) we get p(A[intersect]C)>p(B[intersect]C) similarly for c compliment P(A[interect]c) [union] p(A[interect]C[compliment]) just makes up the whole of p(A) same for B result then follows. Link to comment Share on other sites More sharing options...
jambo_ED Posted October 22, 2008 Share Posted October 22, 2008 37. You're welcome isn't it 42? Link to comment Share on other sites More sharing options...
kevmacd Posted October 22, 2008 Share Posted October 22, 2008 St Johnstone.. Link to comment Share on other sites More sharing options...
hughesie27 Posted October 22, 2008 Share Posted October 22, 2008 Yer Ma Link to comment Share on other sites More sharing options...
Romanov Stole My Pension Posted October 22, 2008 Author Share Posted October 22, 2008 p(A|C) = p(A[intersect]C)/p©p(B|C) = p(B[intersect]C)/p© similarly for c compliment Express inequalities in that form. Cancel the fractions ( muliply by p© or p(C[compliment) ) we get p(A[intersect]C)>p(B[intersect]C) similarly for c compliment P(A[interect]c) [union] p(A[interect]C[compliment]) just makes up the whole of p(A) same for B result then follows. Thank you very much sir. Cheers to Ed as well Link to comment Share on other sites More sharing options...
northfieldhearts Posted October 22, 2008 Share Posted October 22, 2008 Long shot but hopefully some statistics students read this.Question is... Let A, B and C be events. Show that if P(A|C)>P(B|C) and P(A|C[complement])>P(B|C[complement]), then P(A)>P(. If A = YMCA then B = DISCO & C = DIVORCE Link to comment Share on other sites More sharing options...
Seats Posted October 22, 2008 Share Posted October 22, 2008 isn't it 42? ah pants, you're right - I obviously forgot to factor in a(function_p)/c(hobo) It is clearly 42 - hope you are happy now Link to comment Share on other sites More sharing options...
Walter Payton Posted October 22, 2008 Share Posted October 22, 2008 Thank you very much sir.Cheers to Ed as well You've got the answer then? I was just about to PM you:cool: Link to comment Share on other sites More sharing options...
Say What Again Posted October 22, 2008 Share Posted October 22, 2008 You've got the answer then? I was just about to PM you:cool: I think Douglas Adams may have helped provide the answer Link to comment Share on other sites More sharing options...
Romanov Stole My Pension Posted October 23, 2008 Author Share Posted October 23, 2008 You've got the answer then? I was just about to PM you:cool: Nup not yet, I was gonna try to work it out from CommanderHarris's help above but if you could PM me that would be fantastic. Link to comment Share on other sites More sharing options...
Heartbeat Posted October 23, 2008 Share Posted October 23, 2008 A,B,C = 1,2,3 its easy as Do, Ray, Me Answer is simple THE JACKSON 5 Link to comment Share on other sites More sharing options...
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