Can probability of getting a ball from a box be greater than $1$ 
A box has three black balls and three red balls. Therefore, probability of 
getting a black ball is $\dfrac{1}{2}$ (so is the case for red ball
  also)
Suppose one person takes a ball and put it back in the box. Then
  second person takes a ball and put it back in the box. Then third
  person takes a ball and put it back in the box.
What is the probability that at least one person gets a black ball?
(note: question is created myself to clear a doubt)

Probability of first person getting a black ball is $\dfrac{1}{2}$
Probability of second person getting a black ball is $\dfrac{1}{2}$
These three are independent events. So, it appears as $\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}$  is the probability that at least one person gets a black ball. I know this is wrong as I studied in my book that probability cannot be greater than $1$
Then, I myself got this contradictory argument also.

probability that all get black ball
  $=\left(\dfrac{1}{2}\right)^3=\dfrac{1}{8}$
Probability that exactly two of them get a black ball
  $=\dbinom{3}{2}\left(\dfrac{1}{2}\right)^3=\dfrac{3}{8}$
Probability that exactly one of them get a black ball
  $=\dbinom{3}{1}\left(\dfrac{1}{2}\right)^3=\dfrac{3}{8}$
In this way,  Probability that at least one of them get a black ball
  $=\dfrac{1}{8}+\dfrac{3}{8}+\dfrac{3}{8}+=\dfrac{7}{8}$

I have one more argument for this.

Probability that none of them get a black ball $=\left(\dfrac{1}{2}\right)^3=\dfrac{1}{8}$
So, Probability that at least one of them get a black ball
  $=1-\dfrac{1}{8}=\dfrac{7}{8}$

Could you please help me to understand which approach is right? And more importantly, the first argument which is also convincing to me, has probability more than $1$ 
 A: The events are independent, but to add probabilities you need them to be disjoint. The general statement is P(A it B)=P(A)+P(B)-P(A and B). The reason you subtract P(A and B) is you counted it in both of the other terms, so you must subtract it once. You can only add two probabilities directly if they can't both happen. That is what is wrong with your first approach.
A: Either of your last two methods would be considered a good way to compute the probability of at least one black ball drawn.
Setting aside for a moment the question why the first method gives a different answer, a more fundamental question is:

Why is it ever correct to add two probabilities together?

Or even more fundamentally:

What does it mean when we say the probability of something is $\frac12$?

There are some subtleties about the second question (see this question for starters), but one way to think of it is a way to measure the portion of possible worlds in which a certain event occurs out of all possible worlds that might be. For instance, we suppose by symmetry that out of all possible worlds, in exactly half of them the first drawn ball is black and in the other half the first ball is red.
Indeed if two events are disjoint, that is, if there is no possible overlap between the possible worlds in which each event occurs, it makes sense to add the probabilities. The first ball drawn must be red or black, never both red and black, so it is correct to add the probability of red to the probability of black: $\frac12+\frac12=1$.
But the possible worlds in which the second ball is black are not disjoint from the worlds   In which the first ball is black. In fact, in half the worlds in which the second ball is black (one quarter of all possible worlds), the first ball is black too. In only $\frac14$ of all possible worlds, the first ball is red but the second is black. Hence when we consider the first two balls, we find that in $\frac12+\frac14=\frac34$ of all possible worlds, one of the first two draws is a black ball. For similar reasons we find that in $\frac78$ of all possible worlds at least one of the three balls drawn is black, in agreement with the second two methods you tried.
There is another way to set up three people drawing balls so that each person has a $\frac12$ chance to draw a black ball: first, someone else secretly and randomly picks one of two boxes, one filled with black balls and the other with red balls, and the each of the three people draws a marble from the box. Then in $\frac12$ of all possible worlds the first person draws a black ball, and in $\frac12$ of all possible worlds the second person draws a black ball, but those are _exactly the same set of possible worlds, so when we consider the probability that at least one of the three balls drawn is black, it is just $\frac12$.
This last paragraph has not much to do with your question except to illustrate how the way we combine probabilities of events depends on our assumptions about the relationships between events--whether they are disjoint, independent, identical, or have some other amount of "overlap" of possibility.
A: In your first (wrong) approach you use the rule:
$\Pr(A\cup B\cup C)=\Pr(A)+\Pr(B)+\Pr(C)\tag1$
However, this rule can only be used if the events $A,B,C$ are mutually exclusive which is not the case here.
A rule that always works is:
$\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)\tag2$
$A,B$ are by definition mutually exclusive if $A\cap B=\varnothing$, and if that is the case then $\Pr(A\cap B)=0$ so that $(2)$ becomes:
$\Pr(A\cup B)=\Pr(A)+\Pr(B)\tag3$
corresponding with $(1)$ applied on $2$ events.
Usage of the general $(2)$ in order to solve the problem would lead to:
$$\Pr(A\cup B\cup C)=$$$$\Pr(A)+\Pr(B)+\Pr(C)-\Pr(A\cap B)-\Pr(A\cap C)-\Pr(B\cap C)+\Pr(A\cap B\cap C=\frac12+\frac12+\frac12-\frac14-\frac14-\frac14+\frac18=\frac78$$
That answer agrees with your (correct) second and (correct and most efficient) third effort.
