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? Options:
A) 3/4  B) 1/4  C) 1/3  D) 7/19  E) 1/7
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 red nor green   = event that the ball drawn is blue.   n(E)=7   P(E)=n(E)/n(S)   =7/21   =1/3   The answer should be 1/3
<result>C</result>
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?