Question 463713
Let X = number of correct answers in a 20-multiple choice questions.
X~Binomial(n=20, p=0.2)
There are 5 choices so the probability of success is 1/5 or 0.2 and the probability of failure is 1-0.2 = 0.8.
Therefore, the pmf of X is {{{p(x) = (matrix(2,1, 20, x))*(0.2)^x*(0.8)^(20-x)}}}
Using the pmf we can now solve problems A,B, and C.

A.{{{P(X<5) = sum( (matrix(2,1, 20, x))*(0.2)^x*(0.8)^(20-x), x = 0, x = 4 )}}} = 0.6296
B. 80% of 20 is 16 so we compute {{{P(X>=16) = sum( (matrix(2,1, 20, x))*(0.2)^x*(0.8)^(20-x), x = 16, x = 20 )}}}, the result is a very small number close to zero.

C. {{{p(18) = (matrix(2,1, 20, 18))*(0.2)^18*(0.8)^2}}}, the result is also a very small number close to zero.