Initial composition of the bag

There are initially:

• Red balls $=4$
• Black balls $=6$
• Total balls $=10$

First draw: split into cases

A ball is drawn first.

Case 1: First ball is red

Probability of this event: $P(R_1)=\frac{4}{10}=\frac{2}{5}$

Since the drawn red ball is returned along with two additional red balls, the new composition becomes:

• Red balls $=4+2=6$
• Black balls $=6$
• Total balls $=12$

So, probability that the second ball is red in this case: $P(R_2\mid R_1)=\frac{6}{12}=\frac{1}{2}$

Case 2: First ball is black

Probability of this event: $P(B_1)=\frac{6}{10}=\frac{3}{5}$

Since the drawn black ball is returned along with two additional black balls, the new composition becomes:

• Red balls $=4$
• Black balls $=6+2=8$
• Total balls $=12$

So, probability that the second ball is red in this case: $P(R_2\mid B_1)=\frac{4}{12}=\frac{1}{3}$

Use total probability

The required probability is: $P(R_2)=P(R_1)P(R_2\mid R_1)+P(B_1)P(R_2\mid B_1)$

Substituting values: $P(R_2)=\frac{2}{5}\cdot \frac{1}{2}+\frac{3}{5}\cdot \frac{1}{3}$

$P(R_2)=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}$

Compare with options

$\frac{2}{5}$ corresponds to Option C.

Verification with stored answer

Stored correct answer is C, which matches our derived answer.