A multiplechoice exam has six problems each of which has thr

A multiple-choice exam has six problems, each of which has three possible answers. What is the probability that John will get four or more correct answers by just guessing?

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

P( X < 4) = P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 6 3 ) * 0.333^3 * ( 1- 0.333 ) ^3 + ( 6 2 ) * 0.333^2 * ( 1- 0.333 ) ^4 + ( 6 1 ) * 0.333^1 * ( 1- 0.333 ) ^5 + ( 6 0 ) * 0.333^0 * ( 1- 0.333 ) ^6   
= 0.9002
P( X > = 4 ) = 1 - P( X < 4) = 0.0998