Question

A student takes a multiple choice quiz with 12 questions and 5 alternatives per question. Unfortunately, this student has no idea what any of the answers are, so he randomly selects an alternative for each question on the exam. Using this “blind guessing” strategy, how likely is the student to get:

(a). exactly 5 answers correct?

(b). exactly 2 answers correct?

(c). 4 or fewer answers correct?

(d). He wanted to get at least 75% on the quiz. How likely is that to happen?

Answer 1

prob(correct) = 1/5
prob(not correct) = 4/5

a) to have exactly 5 correct
= C(12,5) (1/5)^5 (4/5)^7
= appr .05315

b) prob(2 correct)
= C(12,2) (1/5)^2 (4/5)^10
= ...

c) 4 or fewer correct
---> 0 correct + 1 correct + 2 correct + 3 correct + 4 correct

d) to get 75% he needs to have at least 9 correct
so find
9 correct + 10 correct + 11 correct + 12 correct

lots of calculations, follow the method I used in a) and b)