A bag contains 8 red, 6 green and 7 blue balls. Three balls are drawn one by one at random without replacement. Find the probability that the first ball is red, second ball is green and third ball is blue?

 \(\frac{4}{95}\) 

Red = 8

Green = 6

Blue = 7

Total = 21

Reqd. probability = \(\frac{8}{21}\) ×  \(\frac{6}{20}\) ×  \(\frac{7}{19}\) 

                           =  \(\frac{4}{95}\) 

Hence, option (B) is correct.