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?

A. 1/3
B. 3/4
C. 7/19
D. 8/21
E. 9/21

• To select neither red nor green, the probability is 1-(8C1/21C1+6C1/21C1)
  =1-(14/21)
  =1/3
• 13 years agoHelpfull: Yes(6) No(1)
• Total number of balls = (8 + 7 + 6) = 21.
  n(E) = 7.
  P(E) = n(E)/n(s)=7/21
  n(S)=1/3
  
• 13 years agoHelpfull: Yes(4) No(1)
• Answer: (A)1/3
  
  Solution:
  The sample space=21C1=21
  hence we have to select 1 ball from the 21 balls.
  but there are only 7 balls by which the event can occurs.
  so required probabily is :
  number of events/number of sample space
  =7/21
  =1/3
• 13 years agoHelpfull: Yes(3) No(2)
• A. 1/3
  Its 7C1/21C1 = 7/21 = 1/3
• 13 years agoHelpfull: Yes(2) No(0)
• chances of getting blue=7/21 or
  1/3
• 13 years agoHelpfull: Yes(1) No(0)
• since, neither red ball nor green ball is drawn. its mean blue ball is drawn
  
  so required probability of drowning one ball is P
  
  P= (7C1)/(21C1)= 7/21= 1/3
• 10 years agoHelpfull: Yes(1) No(1)
• total ball=21
  if neither red nor green then it will blue
  hencr probability
  7/21
  =1/3
• 10 years agoHelpfull: Yes(1) No(1)
• Answer: Option A
  
  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.
  
  P(E) = n(E) = 7 = 1 .
  n(S) 21 3
  
  
  
  
• 13 years agoHelpfull: Yes(0) No(0)