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Probability

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Can someone explain answer for this question(26)?

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. $\frac{2}{3}$
  2. $\frac{3}{4}$
  3. $\frac{7}{19}$
  4. $\frac{9}{21}$
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Total number of balls = 8 + 7 + 6 = 21

Let E be the event of selecting neither red nor green ball

Then n(E) = 7

Required probability = 7/21 = 1/3
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