SOLUTION: An exam was applied with 10 questions of multiple choice, with four options each, being only one correct. John took the exam without knowing anything, answering the questions at ra

Question 1105419: An exam was applied with 10 questions of multiple choice, with four options each, being only one correct. John took the exam without knowing anything, answering the questions at random.
a) What is the probability of John to solve four questions?
b) What is the probability that John scores more than 50% of the exam?
c) John's mother promised $ 100 if he takes out 9 or 10, or $ 50 if he takes 7 or 8. If he draws below 7, it goes without premium. Calculate the expected gain of John.

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Getting 4 correct is 10C4(1/4)^4(3/4)^6=0.1460.
Scores more than 50% means getting 6 or more right.
Getting 6 right is 210(.25^6)(.75^4)=0.0162; 7 right is 120*.25^7*.75^3=0.003; 8 right 0.0004, 9 right is <0.0001, so the probability is 0.0196
Expected value is X*p(X), so it is $100* probability of 9 or 10+ $50 for probability of 7 or 8.
9 or 10 right is 0.00003 and 10 right is 0.000001, for 0.0000031, the probability for $100=0.000095
for 7 and 8, the probability is 0.0034. Multiply by $50 to get $0.17. That is the answer given the results for 9 and 10.

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