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OpenStudy (anonymous):

A professor gives a true-false examination consisting of thirty T-F questions. The questions whose answers are "true" are randomly distributed among the thirty questions. The professor thinks that 3/aof the class are serious, and have correctly mastered the material, and that the probability of a correct answer on any question from such students is 75%. The remaining students will answer at random. She glances at a couple of questions from a test picked haphazardly. Both questions are answered correctly. What is the probability that this is the test of a serious student?

OpenStudy (anonymous):

what is 75% of 30 agian (dont have calculator of hand)

OpenStudy (anonymous):

22.5

OpenStudy (anonymous):

I am not sure sorry. But you could say that it is possible that the test is from a "serious student" because you are asked for the possibility which means that it doesnt have to be a percentage. hope this helps

OpenStudy (kropot72):

@7mada2010 There is a typo in this part of your question; "The professor thinks that 3/aof the class are serious". Please check the original question and post a correction.

OpenStudy (usukidoll):

3 out of the class?

OpenStudy (anonymous):

oh i'm sorry i mean 3/4 class

OpenStudy (usukidoll):

75%

OpenStudy (anonymous):

A professor gives a true-false examination consisting of thirty T-F questions. The questions whose answers are "true" are randomly distributed among the thirty questions. The professor thinks that 3/4 of the class are serious, and have correctly mastered the material, and that the probability of a correct answer on any question from such students is 75%. The remaining students will answer at random. She glances at a couple of questions from a test picked haphazardly. Both questions are answered correctly. What is the probability that this is the test of a serious student?

OpenStudy (anonymous):

A: serious students, Ac: remaining students. Q1, Q2: each of the questions 1,2 is answered correctly. P(Q1|A)=P(Q2|A)=0.75. If independence is assumed between the questions, then P(Q1Q2|A)=P(Q1|A)P(Q2|A)=... P(A)=0.75 => P(Ac)=... Since the remaining students choose their answers randomly (i.e either right or wrong), P(Q1|Ac)=P(Q2|Ac)=... Again, we assume independence between 2 questions, then P(Q1Q2|Ac)=P(Q1|Ac)*P(Q2|Ac)=... Now, P(Q1Q2)=P(Q1Q2|A)*P(A)+P(Q1Q2|Ac)*P(Ac)=... The desired probability is: P(A|Q1Q2)=P(Q1Q2|A)*P(A)/P(Q1Q2)=... Now it's your job to fill in all the blanks (correctly, of course!)

OpenStudy (anonymous):

thanks for your help

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