A bag contains 5 Blue and 4 Black balls.
3 balls are drawn at random. What is the probability that 2 are blue and 1 is black?


Approach (1):- Total number of ways of drawing 3 balls from 9 balls
9C3 = 84

No. of ways of drawing 2 Blue from among 5 balls
5C2 = 10

No. of ways of drawing 1 Balck ball from among 4 balls
4C1 = 4

Therefore no. of ways of drawing 2 Blue and 1 Black balls = 14/84 = 1/6 <--- Correct answer





Approach (2):- No. of ways of drawing 2 blue balls:-
5C2/9C2 = 5/18

No. of ways of drawing 1 black ball from remaining 7 balls:-
4C1/7C1 = 4/7

Total no. of ways of drawing 2 Blue AND 1 Black ball = 5/18 * 4/7 = 10/63 <--- Incorrect answer



Please let me know what is wrong with the 2nd approach?
When to apply approach (1) and when to apply approach (2)?