Assume that two students must take a ten question, True or False quiz. Student 1

Question

Assume that two students must take a ten question, True or False quiz. Student 1 has not studied at all and guesses randomly on each question. Student 2 has studied hard and has 95% chance of answering any particluar question correctly.

For each student, cacluate the probablity of passing (getting 7 or more correct answers) and the probablity of "acing" (getting 9 or 10 correct answers) the quiz.

What do your caculated pobablitites say about the amount of prepartaion students do and thier proability of doing well on the quiz?

Explanation / Answer

Ans:

For student guesses:

P(7 or more correct answers)=10C7 0.57 *0.53 +10C8 0.58 *0.52 +10C9 0.59 *0.51 +10C10 0.510 *0.50

=0.0009766*(120+45+10+1)=0.1719

P(9 or 10 correct answers)=10C9 0.59 *0.51 +10C10 0.510 *0.50   =0.0009766*(10+1)=0.0107

For student who has studied hard:

P(7 or more correct answers)=10C7 0.957 *0.053 +10C8 0.958 *0.052 +10C9 0.959*0.051+10C10 0.9510 *0.050

=(120*0.00008729+45*0.001658+10*0.0315+1*0.5987)=0.9988

P(9 or 10 correct answers)=10C9 0.959*0.051+10C10 0.9510 *0.050   =(10*0.0315+1*0.5987)=0.9137

Probability of the student who has studied hard of doing well is very higher than the student who guesses(in both cases)

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