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 nCkP^kQ^(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.