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
1) 1/3
2) 3/4
3) 7/19
4) 8/21
5) 9/21

• suppose side of main square = x
  area of square =x^2
  
  Side of middle square = (x/2)*root 2
  area of inner square = ((x/2)*root 2)^2 = 2*x/4= x^2/2
  
  hence area of inner square is half of main (outer square).
• 12 years agoHelpfull: Yes(18) No(5)
• if side of main square = 2a
  area of square = 4*a^2
  
  Side of middle square = a*root 2
  area of inner square = (a*root 2)^2 = 2*a^2
  
  so area of inner square is half of main (outer square).
• 12 years agoHelpfull: Yes(16) No(3)
• Ans. (1)=1/3
  P=Blue/Total
  = 7/21
  =1/3
• 8 years agoHelpfull: Yes(16) No(0)
• Total number of balls = (8 + 7 + 6) = 21.
  
  Let = 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 .
  
  
  
• 7 years agoHelpfull: Yes(3) No(0)
• it is option 1. 1/3. instead of finding neither red nor green if we find probability for blue that would be the answer.
• 7 years agoHelpfull: Yes(3) No(0)
• 252/900
  19*8 +100 no of 7s =252
  Total no = 900
• 5 years agoHelpfull: Yes(0) No(0)