A bag contains 5 red and 7 black balls. Three balls are chosen at random one by one without replacement. What is the probability that the third ball is a Red ball given that first ball is red and second ball is black?

$\frac{7}{66}$

Probability of getting red ball in first draw = $\frac{5}{12}$

Probability of getting black ball in second draw = $\frac{7}{11}$

Probability of getting red ball in third draw = $\frac{4}{10}= \frac{2}{5}$

Required probability = $\frac{5}{12} \times \frac{7}{11}\times\frac{2}{5} =  \frac{7}{66}$