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?
Answer & explanation
Correct answer: option 1
Orange balls= 9
Green balls= 8
Blue balls= 4
Total Balls = 21
Required Probability = \(\frac{9}{21}\) × \(\frac{8}{20}\) × \(\frac{4}{19}\)
= \(\frac{24}{665}\)