A bag contains 5 red, 4 blue and M green
balls if the probability of getting both the balls
green, when two balls are selected at
random is 1/7, find M

• we have to select 2 balls and it will be green so we will select two balls from green balls
  so
  
  mC2/m+9C2=1/7
  on solving
  
  (m! / (m-2! * 2!) ) / (m+9! / (m+7! * 2!) ) =1/7 From nCr=n! / r!*(n-r)!
  on solving
  (m! / (m-2)! ) * ( (n-7)! / (n+7)!) =1/7
  
  after solving and cross multiplication
  m^2+4m-12=0
  on solving
  m=2,-6
  so m will be 2.. Ans
  
• 8 years agoHelpfull: Yes(0) No(2)