tag:blogger.com,1999:blog-4115025577315673827.post3354023053494195899..comments2021-01-26T19:30:37.111+05:30Comments on CSE Blog - quant, math, computer science puzzles: Fraction BrainteaserPratik Poddarhttp://www.blogger.com/profile/11577606981573330954noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-4115025577315673827.post-48888792506572710892018-08-07T10:20:34.655+05:302018-08-07T10:20:34.655+05:30Although, in both concepts, Siddhant is better tha...Although, in both concepts, Siddhant is better than vaibhav. But see that, both are more good in arithmetic than the geometry. Means if, more arithmetic questions are put, the percentage of total score will go up. Thus for vaibhav due to more questions from arithmetic, he has higher total score.Mukul Bansalhttps://www.blogger.com/profile/06316809584595094231noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-27728115539577409182018-01-24T10:13:35.801+05:302018-01-24T10:13:35.801+05:30Because a/b+c/d != (a+c)/(b+d)Because a/b+c/d != (a+c)/(b+d)Anonymoushttps://www.blogger.com/profile/08688934077375884948noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-74627972469852414702018-01-23T19:41:31.314+05:302018-01-23T19:41:31.314+05:30Let C denote solving problem correctly. Let S deno...Let C denote solving problem correctly. Let S denote solving by Siddhant and A denote Arithmetic problem.<br />Then according to problem P(C | S, A) > P(C | ~S, A) and P(C | S, ~A) > P(C | ~S, ~A) but P(C | S) < P(C | ~S).<br />This problem is very prominent in statistics. Here the variable A is called the confounding variable, it influences both the independent (S) and dependent variable(C). More can be understood from wiki link on "Confounding".random123https://www.blogger.com/profile/09311351499983373506noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-54497113178179431992018-01-02T22:11:03.325+05:302018-01-02T22:11:03.325+05:30each question of Arithmetic =x, Geometry =y (say)...each question of Arithmetic =x, Geometry =y (say)<br />6x+28y=34, 25x+9y=34<br />=> x=y=1<br />Therefore, each has same weightage of 1 mark. <br />Even though, 5/6 > 20/25 and 20/28 > 6/9 Vaibhav answered more (20+6=26) questions than Siddhant(5+20=25)sumanthhttps://www.blogger.com/profile/00752341279310790809noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-24606442556040558012017-12-15T14:30:35.734+05:302017-12-15T14:30:35.734+05:30Score is just the sum of the correct answers. It h...Score is just the sum of the correct answers. It has nothing to do with fractions.Anonymoushttps://www.blogger.com/profile/00094235259094340406noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-23898226561535328272017-12-15T13:41:35.899+05:302017-12-15T13:41:35.899+05:30This is Simpson's paradox. Siddhant is better ...This is Simpson's paradox. Siddhant is better than Vaibhav in each individual subject, but Siddhant's total score is lesser because he attempts more geometry questions, which are harder than arithmetic questions.<br /><br />See https://en.wikipedia.org/wiki/Simpson%27s_paradox for a more detailed explanation.GoKuhttps://www.blogger.com/profile/05674743004147477362noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-5773461392526594202017-12-15T12:55:45.442+05:302017-12-15T12:55:45.442+05:30Score = accuracy * total score
So, for Siddhant hi...Score = accuracy * total score<br />So, for Siddhant higher accuracy (5/6) has lower scoring (6) where as Vaibhav has higher score in more accurate tests(arithmetic) mbj111https://www.blogger.com/profile/08175727921626605252noreply@blogger.com