In a group of 5 students, 2 are males and 3 are females. Two students are
randomly selected (without replacement). Let X be the number of males in the two selected
students.
(a) Find the (probability) distribution of X (i.e., list all possible values of X and their corresponding
probabilities).
(b) Find the expected value of X, and the standard deviation of X.

let 
n = 2 = number of trials
P = 2/5 = probability of success
Q = 3/5 = probability of failure
k = number of success

For a) I used the equation nCk*P^k*Q^(n-k), and got
P(X=0)= 0.36
P(X=1)= 0.48
P(X=2)= 0.16

but the solution key tells me that
P(X=0)= 0.3
P(X=1)= 0.6
P(X=2)= 0.1

And for b)
E(X)= np = 0.8
SD = sqrt(npq) = 0.48

but the solution key tells me that 
E(X) = 0.8
SD = 0.6


And now i'm confused... 
I'm not sure if I should approach this question using the binomial probability or is it completely unrelated to binomial distribution.
Please give me a hint on how to approach this question, thank you.