What is the probability he got at least 2 out of 10 questions correct having 4 answer choices?

0.756



Explanation:

We have two possible events A and B
A = Stan gets 2 or more questions correct
B = Stan gets none correct, or Stan gets exactly one correct

Finding P(A) is a lot of work doing so directly without using P(B). Why? Because we need to find P(2), P(3), P(4), ... all the way up to P(10). Then we add up all those results. The notation P(x) means the probability of getting x correct. 

If we find P(B) first, then we can use the fact that
P(A) + P(B) = 1
P(A) = 1 - P(B)
which is a more efficient route.
Note how A and B are complementary events


Let's find P(0) which is the probability of getting x = 0 correct
n = 10
x = 0
n C x = 10 C 0 = (10!)/(0!*(10-0)!) = 1
P(x) = (n C x)*(0.25)^x*(1-0.25)^(n-x)
P(x) = (n C x)*(0.25)^x*(0.75)^(n-x)
P(x) = (10 C x)*(0.25)^x*(0.75)^(10-x)
P(0) = (10 C 0)*(0.25)^0*(0.75)^(10-0)
P(0) = (1)*(0.25)^0*(0.75)^10
P(0) = 0.05631351470948

Repeat for x = 1 as well
n = 10
x = 1
n C x = 10 C 1 = (10!)/(1!*(10-1)!) = 10
P(x) = (n C x)*(0.25)^x*(1-0.25)^(n-x)
P(x) = (n C x)*(0.25)^x*(0.75)^(n-x)
P(x) = (10 C x)*(0.25)^x*(0.75)^(10-x)
P(1) = (10 C 1)*(0.25)^1*(0.75)^(10-1)
P(1) = (10)*(0.25)^1*(0.75)^9
P(1) = 0.18771171569824

So we see that,
P(B) = getting 0 correct or getting 1 correct
P(B) = P(0) + P(1)
P(B) = 0.05631351470948+0.18771171569824
P(B) = 0.24402523040772



Now we can find P(A)
P(A) = probability of getting at least two correct
P(A) = 1 - P(B)
P(A) = 1 - 0.24402523040772
P(A) = 0.75597476959229
P(A) = 0.756
The probability of getting at least two correct is roughly 0.756