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2-students-are-passing-a-test-of-n-questions-with-the-same-chance-to-find-each-one-Show-the-chance-that-they-both-don-t-find-a-same-question-is-3-4-n-




Question Number 207372 by sniper237 last updated on 12/May/24
2 students are passing   a test of  n questions with  the same chance to find each one  Show  the chance that they both   don′t find a same question is  ((3/4))^n
2studentsarepassingatestofnquestionswiththesamechancetofindeachoneShowthechancethattheybothdontfindasamequestionis(34)n
Commented by A5T last updated on 13/May/24
If the chance of finding(getting ?) one question   is (x/y)≤1;and the students be A and B. When n=1;   there are three cases:  (i) A gets the question and B misses.  (ii) B gets the question and A misses  (iii) A and B miss the question  ⇒Probability=(x/y)(1−(x/y))+(1−(x/y))((x/y))+(((y−x)^2 )/y^2 )  =((x(y−x))/y^2 )+(((y−x)^2 )/y^2 )=(((y−x)(y−x+x))/y^2 )=((y−x)/y)  So, the probability is dependent on (x/y).   For example, if (x/y)=(1/4); then probability=(3/4).  But if (x/y)=1; then probability=0
Ifthechanceoffinding(getting?)onequestionisxy1;andthestudentsbeAandB.Whenn=1;therearethreecases:(i)AgetsthequestionandBmisses.(ii)BgetsthequestionandAmisses(iii)AandBmissthequestionProbability=xy(1xy)+(1xy)(xy)+(yx)2y2=x(yx)y2+(yx)2y2=(yx)(yx+x)y2=yxySo,theprobabilityisdependentonxy.Forexample,ifxy=14;thenprobability=34.Butifxy=1;thenprobability=0
Commented by sniper237 last updated on 13/May/24
Hint: The results of that experience   are  sets {(A, B)/ A,B⊂{1,....,n}}   of questions correctly  answered  And the event here are all those  such as A∩B=∅
Hint:Theresultsofthatexperiencearesets{(A,B)/A,B{1,.,n}}ofquestionscorrectlyansweredAndtheeventhereareallthosesuchasAB=
Commented by A5T last updated on 13/May/24
It is dependent on  (x/y)(the probability of them  getting the correct answer). So,unless (x/y)=(1/4),  ((3/4))^n  failed to hold for n=1.
Itisdependentonxy(theprobabilityofthemgettingthecorrectanswer).So,unlessxy=14,(34)nfailedtoholdforn=1.
Commented by sniper237 last updated on 13/May/24
Read the post again. you don′t need that
Readthepostagain.youdontneedthat
Commented by A5T last updated on 13/May/24
Do you have a solution?
Doyouhaveasolution?
Commented by A5T last updated on 13/May/24
(i) A gets the question and B misses.  (ii) B gets the question and A misses  (iii) A and B miss the question  (iv) A and B get the question    Then there are (3/4) ways for A and B not to get  the same question?  So, ((3/4))^n  for n questions?
(i)AgetsthequestionandBmisses.(ii)BgetsthequestionandAmisses(iii)AandBmissthequestion(iv)AandBgetthequestionThenthereare34waysforAandBnottogetthesamequestion?So,(34)nfornquestions?

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