A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its color is observed and this ball along with two additional balls of the same color are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is:

First, we note the contents of the bag before any drawing takes place. There are $$4$$ red balls and $$6$$ black balls, giving a total of $$10$$ balls.

We draw one ball at random and observe its colour. Let us compute the probability of each possible colour for this first draw. Since all balls are equally likely to be chosen, we have

$$P(\text{first ball is red})=\frac{4}{10}=\frac{2}{5},\qquad P(\text{first ball is black})=\frac{6}{10}=\frac{3}{5}.$$

After seeing the colour, we return that same ball and also place two extra balls of the same colour back into the bag. We now follow both cases separately in complete detail.

Case 1 - The first ball is red.

We removed one red, so the reds momentarily drop to $$3$$, but we immediately return the original red along with two more red balls. Thus the count of red balls becomes

$$3+3=6.$$

The black balls remain unchanged at $$6.$$ Hence the total number of balls in the bag is now

$$6+6=12.$$

The probability that the second (new) draw gives a red ball under this condition is therefore

$$P(\text{second red}\mid\text{first red})=\frac{6}{12}=\frac{1}{2}.$$

Case 2 - The first ball is black.

We removed one black, leaving $$5$$ black balls. Putting the original black back along with two more blacks adds $$3$$ blacks, so the black count increases to

$$5+3=8.$$

The number of red balls stays at $$4.$$ Thus the total number of balls is again

$$4+8=12.$$

The probability that the second draw gives a red ball in this situation is

$$P(\text{second red}\mid\text{first black})=\frac{4}{12}=\frac{1}{3}.$$

Now we combine these two conditional probabilities using the Law of Total Probability, which states

$$P(A)=P(B_1)\,P(A\mid B_1)+P(B_2)\,P(A\mid B_2)$$

when $$\{B_1,B_2\}$$ form a partition of the sample space. Here, $$A$$ is “second ball is red”, $$B_1$$ is “first ball is red”, and $$B_2$$ is “first ball is black”. Substituting the numbers we have already calculated,

$$\begin{aligned} P(\text{second red}) &=P(\text{first red})\,P(\text{second red}\mid\text{first red}) +P(\text{first black})\,P(\text{second red}\mid\text{first black})\\[6pt] &=\left(\frac{2}{5}\right)\left(\frac{1}{2}\right) +\left(\frac{3}{5}\right)\left(\frac{1}{3}\right)\\[6pt] &=\frac{2}{5}\cdot\frac{1}{2}+\frac{3}{5}\cdot\frac{1}{3}\\[6pt] &=\frac{1}{5}+\frac{1}{5}\\[6pt] &=\frac{2}{5}. \end{aligned}$$

Thus the probability that the ball drawn in the second draw is red equals $$\frac{2}{5}.$$

Hence, the correct answer is Option C.