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
8/11
2/11
4/15
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.
n(E) = 7.
$\therefore$ P(E) = n(E)/n(S) = 7/21 = 1/3.