A multiple choice examination has 5 questions, each question has three alternative
answers of which exactly one is correct. If p is the probability that a student will get 4 or
more correct answers just by guessing, then 34p=
n(s)=35A⇒4 Or more correct ⇒n(A)=5c4×2+5c5=5×2+1=11P(A)=1135=p34p=113=3.67