Prep for USMLE 
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Author  4 Posts  
ali2010
Forum Newbie Topics: 9 Posts: 26 
I am sorry but i am writing here because no epidemiology category q1 in is pg of boistatistics chapter in Kaplan psychiatric note book if the event r nonindependent.....e,g if one has a box with 5 white & 5 blacks balls in it,the chance of picking two blacks balls is (5/10)x(4/9)=0.5x0.44=0.22(or 22%) can u plz explain from where that equation comes from !!!!!!!!!!!!!!!!! q2 in Kaplan epidemiology video when he explain the equation if fat=30% &DN=10% calculate the chance of meet fat or DM or both he said (A+B)(AxB) BUT if ask chance of meet fat or DM not both he said (A+B)2AB BUT i think the equation should be (A+B)2(A/B) correct me if I am wrong q3 can u explain the answer of the last q (q6 ) in Kaplan note 20082009 which is article with tabular data of relative risk of DM the answer was c which is :Level 3 and 4 are significantly different from baseline and risk elevation> why not a,b,d,e again I am sorry but iam writing here because no epidemiology category  
alirizvi
Forum Newbie Topics: 36 Posts: 542 
is this for step 2 ?  
ali2010
Forum Newbie Topics: 9 Posts: 26 
yes, step 2 from kaplan note of psychatric & epidemiology20082009 in chapter 2 biostatistic page:141 combine propapility in is pg of boistatistics chapter in Kaplan psychiatric note book if the event r nonindependent.....e,g if one has a box with 5 white & 5 blacks balls in it,the chance of picking two blacks balls is (5/10)x(4/9)=0.5x0.44=0.22(or 22%) can u plz explain from where that equation comes from !!!!!!!!!!!!!!!!!  in biostatistic page:155 the last q can u explain the answer of the last q (q6 ) in Kaplan note 20082009 which is article with tabular data of relative risk of DM the answer was c which is :Level 3 and 4 are significantly different from baseline and risk elevation> why not a,b,d,e again I am sorry but iam writing here because no epidemiology category the whole note is easy except those  
alirizvi
Forum Newbie Topics: 36 Posts: 542 
omg i hate math, i thought i was done with this stuff .. anyways have no idea and dnt have the kaplan books but what i remember from step 1 is that for nonindependent ( dependent ) event u assume that one has already occured. so if there were 10 balls 5 white and 5 black the chance of u picked black is 5/10 but if u picked one already and one to pick again so 4/9 and the chance of white still is 5/10 since u can still pick those. they basically depend on each other thats the main thing. the other thing i just look at my old f.a book for step 1 and i have a+b  2(axb) .. i dnt think it can be division because when u divide fractions they actually increase like .30/.10 = 3 vs .30 x .10 = .03 i mean the chance of finding someone with both and not both should be harder than finding with one thing. i dnt know if that makes sense. but anyways good luck man  