A bag contains 9 orange balls, 8 green and 4 blue balls. Three balls are drawn one by one at random without replacement. Find the probability that the first ball is orange , second ball is green and third ball is blue?

\(\frac{24}{665}\)

Orange balls= 9 

Green balls= 8

Blue balls= 4

 Total Balls = 21

Required Probability =  \(\frac{9}{21}\) ×   \(\frac{8}{20}\) ×   \(\frac{4}{19}\)

                              =  \(\frac{24}{665}\)