Mykill Posted October 27, 2011 Report Share Posted October 27, 2011 now ur suggesting the probability of picking a card marked X which becomes 16/52now lets do this, put an x on all diamonds, what is the probability of picking card marked x = 13/52 now put a Y on all aces probability is 4/52 what is the probability that of picking either x or y ... i pick back to back 16 x put it back then back to back 4 y then it is 17, the condition never excluded a thing with both x and y.\EditBy far the most hated class I have ever had to take. Quote Link to comment Share on other sites More sharing options...
jpeshek32 Posted October 27, 2011 Report Share Posted October 27, 2011 The only correct answer is 50%. You either will or you won't. Stats is easy. Quote Link to comment Share on other sites More sharing options...
magley64 Posted October 27, 2011 Author Report Share Posted October 27, 2011 Prof just caved when I offered up the x replacement Quote Link to comment Share on other sites More sharing options...
chevysoldier Posted October 27, 2011 Report Share Posted October 27, 2011 288/2./thread Quote Link to comment Share on other sites More sharing options...
fizzer Posted October 27, 2011 Report Share Posted October 27, 2011 The answer is def. 16/52Doc, try to use your logic to answer this slight modification: what are the odds of drawing a red suit, a diamond, or a heart? Clearly there are overlapping sets here, but you cannot say that 26/52(red) + 13/52(diamond) + 13/52(heart) = 52/52, otherwise you would have a 100% probability of drawing a red suit, diamond or a heart. You have to reduce cards that are included twice. Quote Link to comment Share on other sites More sharing options...
NinjaDoc Posted October 27, 2011 Report Share Posted October 27, 2011 The answer is def. 16/52Doc, try to use your logic to answer this slight modification: what are the odds of drawing a red suit, a diamond, or a heart? Clearly there are overlapping sets here, but you cannot say that 26/52(red) + 13/52(diamond) + 13/52(heart) = 52/52, otherwise you would have a 100% probability of drawing a red suit, diamond or a heart. You have to reduce cards that are included twice.in your example the criteria is red, there is no confusion there, ie a card which is red which can be either diamond or heartCard >> RED ( 1 probability) >> diamond or heartbut the question above there is one card which satisfies both criteria, but are mutually exclusive. here a card which is either ace or diamond.Card>> ace (1 probability) or diamond (1 probability) Then again i am just stating a different point of view, am not saying i am right. Quote Link to comment Share on other sites More sharing options...
NinjaDoc Posted October 27, 2011 Report Share Posted October 27, 2011 may be i am basing my argument more on psycho-philosophical base rather than maths i mean i am setting this argument from the observer basis saying that he is ignoring the fact that ace of diamond that he drew twice is one and the same card. Quote Link to comment Share on other sites More sharing options...
NinjaDoc Posted October 27, 2011 Report Share Posted October 27, 2011 (edited) am sorry now i get it , we are choosing blind... in the end it was as straight forward as it could be..dayum Edited October 27, 2011 by NinjaDoc Quote Link to comment Share on other sites More sharing options...
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