What is the probability of getting a blue ball if you pick a random colored ball from a bag? Let's say I have a bag of 10 balls all different color (One of them is blue).
I want to calculate the probability of getting blue if I randomly draw it from a bag 3 times. 
There are two scenarios
Without replacement
1/10 * 9/9 * 8/8 * 3 = 3/10 
With replacement
1 - P(Not getting blue) = 1 - 0.9^3 = 0.271
The thing that confuses me is why can't I think of the REPLACEMENT case in this way
P(Getting blue for 1st draw) + P(Getting blue for 2nd draw) + P(Getting blue for 3rd draw) = 1/10 + 1/10 + 1/10 = 3/10?
Now let's assume you win 10 dollars for every blue ball you get, without replacement you can win maximum of 10 dollars but with replacement you can win maximum of 30 dollars.
According to expected value,
Without replacement
3/10 * 10 = 0.3
With replacement
0.271 * 10 = 0.271
So the game without replacement have better expected value? I feel like I am misunderstanding some important concepts.
1) What is the correct probability of getting blue ball with replacement if you can win multiple times? Can I use binomial probability for this?
2) What is the expected winning of the game with replacement if you can win multiple times?
 A: With regard to how much money you can win by drawing the blue ball, assuming you earn $10$ dollars every time you draw the blue ball, the scenario without replacement has only two possible outcomes:
$0$ blue balls, probability $P(X=0) = 0.7,$ gain $0.$
$1$ blue ball, probability $P(X=1) = 0.3,$ gain $10.$
The expected value of your winnings is
$$ E(10X) = 0(0.7) + 10(0.3) = 0 + 3 = 3.$$
(Note that $\frac3{10}\times 10 = 0.3 \times 10 \neq 0.3.$ Be careful what you write!)
The scenario with replacement has these possible outcomes:
$0$ blue balls, probability $P(Y=0) = \left(\frac9{10}\right)^3 = 0.729,$ gain $0.$
$1$ blue ball, probability 
$P(Y=1) = \binom31 \left(\frac1{10}\right) \left(\frac9{10}\right)^2 = 0.243,$
gain $10.$
$2$ blue balls, probability 
$P(Y=2) = \binom32 \left(\frac1{10}\right)^2 \left(\frac9{10}\right) = 0.027,$
gain $20.$
$3$ blue balls, probability $P(Y=3) = \left(\frac1{10}\right)^3 = 0.001,$
gain $30.$
Note that if we just care about whether we draw blue ball at least once, we can find the probability whether that happens either by adding up all the cases where it does happen,
$$P(Y=1) + P(Y=2) + P(Y=3) = 0.243+0.027+0.001 = 0.271,$$
or by taking $1$ and subtracting the probability that it does not happen,
$$1 - P(Y=0) = 1 - 0.729 = 0.271.$$
But when you gain $10$ for each time you draw the blue ball, the expected value of your winnings is
\begin{align}
E(10Y) &= 0(P(0)) +  10(P(1)) +  20(P(2)) +  30(P(3)) \\
&= 0 + 10(0.243) + 20(0.027) + 30(0.001) \\
&= 0 + 2.43 + 0.54 + 0.03 \\
&= 3.
\end{align}
In fact the expected number of blue balls drawn is exactly the same with or without replacement,
and the expected payment (receiving $10$ dollars each time blue is drawn) also is exactly the same in each case.
Most people would interpret "the probability of getting blue" in each scenario as the probability of getting blue at least once.
This probability is less in the scenario without replacement
(or to put it another way, the probability of not even once drawing blue is greater), 
but once you start counting the number of times blue is drawn (and paying $10$ dollars each time)
the chance of getting a double or triple payout makes up for the increased chance of getting no payout.
When you ask about a probability, you are asking about something that either happens or does not happen, two possible outcomes.
It is possible to set up an expected value that also deals with only two possible outcomes, but expected value very often deals with more than two possible outcomes (in your example with replacement, four possible outcomes), and in those cases it tends to give answers different than you would get by looking only at two possibilities.
A: Yes, the game without replacement have bave better expected value, as with drawing not-blue balls the probability of drawing blue gets higher.
Hint
In with replacement scenario, once again try to calculate probability of the opposite event.
1) If you mean the probability of getting at least one blue ball, then $0.271$. You can use binomial.
2) If you mean expected number of blue ball drawed it is $ 3 * \frac{1}{10} * \frac{9}{10}*\frac{9}{10}+ 3*2* \frac{1}{10} * \frac{1}{10}  * \frac{9}{10} + 3*  (\frac{1}{10})^3 $
