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)?
Probability question
This topic has expert replies
I think my 2nd approach is incorrect because I considered picking up 2 blue balls and then 1 black ball...
The 2nd approach should have been what clock60 has suggested.. but even that doesn't match 10/63!
Any ideas why there are 2 different answers via the 2 approaches?
The 2nd approach should have been what clock60 has suggested.. but even that doesn't match 10/63!
Any ideas why there are 2 different answers via the 2 approaches?