Balls
Let us have a look at some probability questions based on Balls.
Type 1: Balls drawn without Replacement
Q. A bag contains 4 red, 4 green and 4 blue balls. Two balls are drawn at random without replacement. What is the probability that none of the balls drawn is green?
(a) 7/18 (b) 14/33 (c) 7/36 (d) 2/3
Explanations :
Method I:
Total non-green balls = 8
Hence chance of getting one in first draw = 8/12
After first successful draw, non-green balls = 7. Also, total number of balls is now 11.
Hence, probability of getting non-green ball = 7/11
Final probability = 8/12 × 7/11 = 14/33
Answer: (b)
Method II: Using concept of Combinations
Number of ways to choose two non-green balls = $C^{8}_2$ = 28
Total number of ways to choose 2 balls = $C^{12}_2$ = 66
Probability = 28/66 = 14/33
Answer: (b)
Q. A bag contains 4 red, 4 green and 4 blue balls. Two balls are drawn at random without replacement. What is the probability that one ball is blue and other green?
(a) 1/9 (b) 4/33 (c) 14/33 (d) 8/33
Explanation:
Blue balls = 4, Green balls = 4
Probability = 2 × 4/12 × 4/11 = 8/33
We multiplied by 2 as there are two possibilities – green ball drawn 1st and blue ball drawn 1st.
Answer: (d)