Math, Probability on replacement and with out replacement!! I am weak in probability, I am confused with replacement and without replacement, can someone please explain the problem..

A bag contains 3 blue and 2 red marbles, two marbles are selected at random, what is the probability of getting 2 blue marbles with and without replacing first one..

P.S I can't award the points to the reply because I have no enough credits to do so.....
Thanks for the valuable time....
 A: Without replacement: The probability that the first marble you draw is blue is $\dfrac{3}{3+2}$.  There are now $2$ blue marbles and $2$ red marbles left, so, given that the first is blue, the probability that the second marble you draw is blue is $\dfrac{2}{2+2}$.  So the probability that both are blue is  $\dfrac{3}{3+2} \times  \dfrac{2}{2+2} = \dfrac{3}{10} = 0.3$. 
With replacement: The probability that the first marble you draw is blue is $\dfrac{3}{3+2}$.  You replace that marble so there are now $3$ blue marbles and $2$ red marbles left, so the probability that the second marble you draw is blue is $\dfrac{3}{3+2}$.  So the probability that both are blue is  $\dfrac{3}{3+2} \times \dfrac{3}{3+2} = \dfrac{9}{25} = 0.36$.    
A: The difference between drawing with replacement and without replacement is the sample space and the probabilities you get out of the space.
If you are drawing from a set of objects $X$ with replacement $n$ times, then the sample space is the cartesian product $X^n$.  That is, it's the set of ordered $n$-tuples $(X_1, X_2, \ldots X_n)$, where each $X_i \in X$.  Because you put the object "back" into the bag after you draw it, the probability of drawing the same object on the next drawing is $1/|X|$.  In fact, on every drawing, the probability of drawing any particular object is $1/|X|$.
If you are drawing without replacement $n$ times, the sample space isn't the same.  In particular, the sequence of drawings is an $n$-permutation of elements of $X$.  Since you are not putting an object "back" in after drawing it, the probability of drawing that same object again is $0$.
So, for example, if $X = \{a,b,c\}$, you can draw the sequence $(a,a,a)$ if you are drawing with replacement $3$ times.  But you cannot draw that sequence if you are drawing without replacement.
