An urn contains 5 red and 5 black balls. A ball is drawn at random, its color is noted and is returned to the urn. Moreover, 2 additional balls of the same color are put in the urn and then a ball is drawn at random. The probability that the second drawn ball is red, is:

$\frac{1}{2}$

The correct answer is Option (1) → $\frac{1}{2}$ **

Initial composition: 5 red, 5 black

Probability(first red) = $\frac{5}{10}=\frac{1}{2}$

Probability(first black) = $\frac{1}{2}$

Case 1: First ball is red

Two more red balls added → urn becomes 7 red, 5 black

Probability(second red) = $\frac{7}{12}$

Case 2: First ball is black

Two more black balls added → urn becomes 5 red, 7 black

Probability(second red) = $\frac{5}{12}$

Total probability:

$P(\text{second red})=\frac{1}{2}\cdot\frac{7}{12}+\frac{1}{2}\cdot\frac{5}{12}$

$=\frac{7+5}{24}=\frac{12}{24}=\frac{1}{2}$

The probability that the second drawn ball is red is $\frac{1}{2}$.