6red chips ,7green and 8 blue chips are in a bag if 5 are drawn with replacement .what is the probabaility that atleat 3are red?

• (6*6*6*15*15+6*6*6*6*15+6*6*6*6*6)/21^5
  
• 11 years agoHelpfull: Yes(27) No(5)
• 6c3*15c2+6c4*15+6c5/21c5
• 11 years agoHelpfull: Yes(5) No(2)
• 6c3*15c2+6c4*15c1+6c5
• 11 years agoHelpfull: Yes(4) No(5)
• (6c3*15c2+6c4*15c1+6c5)/21c5
• 11 years agoHelpfull: Yes(4) No(0)
• 6c5+6C4*7+6C4*8+6c3*8c2+6c3*7c2+6c3*7*8
• 11 years agoHelpfull: Yes(2) No(2)
• 5c3(2/7)^3(5/7)^2 + 5c4(2/7)^4(5/7) + 5c5(2/7)^5
• 11 years agoHelpfull: Yes(1) No(0)
• 6c3*15c2/21c5 (is it correct or not )if not why????
• 11 years agoHelpfull: Yes(0) No(3)
• 
  as rrrrr=1
  rrrrg=5 ways
  rrrrb=5 ways
  rrrgg=10 ways(5!/(3!*2!)
  rrrbb=10 ways
  rrrgb=20 ways
  total=51 ways
  there are total combination 21c5/(6!*7!*8!)=x
  ans 51/x
• 11 years agoHelpfull: Yes(0) No(0)
• sry
  x=21c5/(6c5*7c5*8c5)
  51/x ans might be
• 11 years agoHelpfull: Yes(0) No(1)
• (6*6*6*15*15+6*6*6*6*15+6*6*6*6*6)/21^5
• 11 years agoHelpfull: Yes(0) No(1)