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 $1$ 

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.
