In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

1 3
3 4
7 19
8 21

Correct answer: 1 3

Solution

The ball must be blue, and there are 7 blue balls. Total balls = 8 + 7 + 6 = 21. So the probability is \(\frac{7}{21} = \frac{1}{3}\).

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