There are 8 red, 7 blue and 6 green balls, 1 ball is picked up randomly thenwhat is the probability that neither blue nor green?	Options:
A) 2/21
B) 3/31
C) 4/17
D) 3/19
E) probability=8/21
Total number of balls = (8 + 7 + 6)= 
<gadget id="calculator">(8 + 7 + 6)</gadget>
<output>21</output>
21   Let E= event that the ball drawn is neither blue nor green =e vent that the ball drawn is red.   Therefore, n(E) = 8.   P(probability=8/21) = 8/21.
<result>E</result>
There are 8 red, 7 blue and 6 green balls, 1 ball is picked up randomly thenwhat is the probability that neither blue nor green?