prob(right ) = 1/5
prob(wrong) = 4/5
let the number that are right be x
C(100,80) (1/5)^80 (4/5)^20
I am stymied for a method to evaluate this, my calculator cannot handle it
Wolfram says:
http://www.wolframalpha.com/input/?i=evaluate+C%28100%2C80%29+%281%2F5%29%5E80+%284%2F5%29%5E20+
There is a 100 question multiple choice test each with 5 choices. You randomly guess each question.
Set up the binomial probability formula to find the probability of getting exactly 80 correct out of 100.
Check to see if you can use the normal approximation to binomial probabilities. then use normal approximation to compute the probability of getting 80 or more correct.
3 answers
I got that but i am not sure how to do the bottom part. when i use normal approximation to commute it i get 0%
notice the result is
7.47 x 10^-38 which I would consider close to zero.
7.47 x 10^-38 which I would consider close to zero.