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....

• Welcome to math.se! Just so you know, most people on here don't visit for the points. We're all just volunteers who really like math. And we're happy to help people who like math. Or are at least trying to learn it. I understand that this is mostly a conceptual question. If you ever ask a homework question, please be sure to type up what you tried! – nomen Jul 27 '14 at 14:57
• Thanks for positive stream,,, It's not homework, more like working on problems all by myself to make good aptitude skills and get a fast brain response... – Ganesh Vellanki Jul 27 '14 at 16:37

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 $\dfrac{3}{3+2} \times \dfrac{3}{3+2} = \dfrac{9}{25} = 0.36$.
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.