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?
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 :)
To find the probability that the student gets 3 or more questions right, you need to add up the probabilities of getting exactly 3, exactly 4, and exactly 5 questions right.
The probability of getting exactly 3 questions right is calculated by multiplying the probability of getting one question right (0.25) by itself three times since there are three questions: (0.25)^3 = 0.015.
Similarly, the probability of getting exactly 4 questions right is (0.25)^4 = 0.003906.
Finally, the probability of getting exactly 5 questions right is (0.25)^5 = 0.0009765.
Now, to find the overall probability of getting 3 or more questions right, you add up these probabilities:
0.015 + 0.003906 + 0.0009765 = 0.0198825.
Therefore, the overall probability that the student gets 3 or more questions right is approximately 0.0198825, or about 1.99%.