prog(right) = 1/5
prob(wrong) = 4/5
prob of your stated event
is, getting none right --> C(12,0)(1/5)^0 (4/5)^12 = .0687
getting one right --> C(12,1) (1/5) (4/5)^11 = .20616
getting two right --> C12,2) (1/5)^2 (4/5)^10 = .2835
getting three right --> C(12,3) (1/5)^3 (4/5)^9 = .23622
sum of these = appr .7946
btw, since passing is probably getting 6 or more correct, we have to do just two more cases for the failing situation
C(12,4)(1/5)^4(4/5)^8 = .1329
C(12,5) (1/5)^5 (4/5)^7 = .05315
for a total of .9806
Or you would have less than 2% chance of passing by merely guessing
A test consists of 12 multiple choice questions with 5 choices for each question. As an experiment, you GUESS on each and every answer without even reading the questions. Probability of getting less than 4 questions correct on this test is:
1 answer