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 blue nor green?
A
$$\frac{{2}}{{3}}$$
B
$$\frac{{3}}{{7}}$$
C
$$\frac{{8}}{{21}}$$
D
$$\frac{{9}}{{22}}$$
Correct Answer: $$\frac{{8}}{{21}}$$
Total number of balls = (8 + 7 + 6) = 21Let E = event that the ball drawn is neither blue nor green = event that the ball drawn is red.
Therefore, n(E) = 8
P(E) = $$\frac{{8}}{{21}}$$
