Question

There's 5 multiple choice questions on a quiz. Four choices to each question. Find the probability that the student gets 3 or more questions right.

p(x=3) (.25)^3= 0.015
p(x=4) (.25)^4=3.906
p(x=5) (.25)^5=9.765

Do you add them all together to get the overall probability?

Answer 1

first .25^5 = .0009765
NOT what you got

This is a BINOMIAL DISTRIBUTION

P(5,k) = C(5,k)p^k (1-p)^(5-k)
here p = .25 and 1-p = .75

Now the problem
3 right:
C(5,3) = 5!/[3! 2!] = 5*4/2=10
P(5,3) = 10 * .25^3*.75^2
= .0879

4 right
C(5,4) = 5!/[4! 1!] = 5
P(5,4) = 5 * .25^4 * .75^1
= .0146

5 right (fat chance :)
C(5,5) = 5!/[5! 0!] = 1
note 0! is defined as one
P(5,5) = 1 *.25^5 * .75*0
= .00009765 as you know

now add those three :)