Probability on geometry and drawing of balls . 
(i) There are $4$ red and $6$ black balls. A ball is drawn at random, its colour is observed and this ball with another two balls of same colour are returned.  Now, if a ball is drawn at random, what is the probability that the ball is red? 

MY WORK :
If the ball drawn at first is red, then the probability that the last one is red:
$$\frac{6}{12}$$
But, if the ball drawn at first us black, then the probability of the last one to be red:
$$\frac{4}{12}$$
So, the probability is:
$$\frac{6}{12}+\frac{4}{12}$$
$$=\frac{5}{6}$$ ...
But, my answer doesn't match. Why?

(ii) $6$ points are taken inside a circle  .  What is the probability that the points lie in the semi circle? 

MY WORK :
For a particular point, the probability is:
$$\frac{\text{Area of semi circle}}{\text{Area of circle}}$$
$$=\frac{1}{2}$$
So, for $6$ points, the probability becomes:
$$\frac{1}{2^6}$$ 
Am I correct ?

The probability of the second ball being red is the sum of probabilities of choosing a red ball both times and choosing a black ball first then a red ball.
$\frac{4}{10}\cdot \frac{6}{12} + \frac{6}{10}\cdot \frac{4}{12} = \frac{48}{120} = \frac{2}{5}$
Your answer would be correct for the second question if you defined the half of the circle before selecting the points. This isn't the case.
It matters not where the first point is placed and assuming this point is within a specific semi circle, say upper or lower, or left or right, after that its a sequence of probabilities that the remaining five points are within that semicircle. Because the first point can be any one of $6$ the probability is:
$$6\cdot (\frac{1}{2})^5$$
