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)


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