Urn I contains 6 red balls and 4 black balls and Urn II contains 4 red balls and 6 black balls. One ball is drawn at random from Urn I and placed Urn II. If one ball is drawn at random from Urn II, then the probability that it is a red ball is :

$\frac{23}{55}$

The correct answer is option (3) → $\frac{23}{55}$

Case 1: Red ball placed from Urn I → II

then combined probability to chose red ball 

from Urn II → $\frac{6}{10}×\frac{5}{11}=\frac{15}{55}$

Case 2: Black placed from Urn I → II

so probability → $\frac{4}{10}×\frac{4}{11}=\frac{8}{55}$

Total probability = $\frac{15}{55}+\frac{8}{55}=\frac{23}{55}$