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?
$ \dfrac{1}{3} $
$ \dfrac{3}{4} $
$ \dfrac{7}{19} $
$ \dfrac{8}{21} $
Explanation:
Total number of balls = (8 + 7 + 6) = 21.
Let E= event that the ball drawn is neither red nor green
= event that the ball drawn is blue.
$\therefore n \left(E\right)$ = 7.
$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{7}{21} $=$ \dfrac{1}{3} $.