There are 6 red balls,8 blue balls and 7 green balls in a bag. If 5 are drawn with replacement, what is the probability at least three are red?

• At least 3 reds means we get either : 3 red or 4 red or 5 red. And this is a case of replacement.
  case 1 : 3 red balls : 6/21 x 6/21 x 6/21 x 15/21 x 15/21
  case 2 : 4 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 15/21
  case 3 : 5 red balls : 6/21 x 6/21 x 6/21 x 6/21 x 6/21
  
  Total probability = = (6/21 x 6/21 x 6/21 x 15/21 x 15/21)+(6/21 x 6/21 x 6/21 x 6/21 x (15 )/21)+ (6/21 x 6/21 x 6/21 x 6/21 x 6/21)
  = 312/16807
  
• 8 years agoHelpfull: Yes(10) No(0)
• case 1=(r,r,r,b,g)=6/21 * 6/21 * 6/21 * 8/21 *7/21=
  case 2=(r,r,r,r,b)=6/21 * 6/21 *6/21 * 6/21 * 8/21=
  case 3=(r,r,r,r,g)=6/21 * 6/21 * 6/21 * 6/21 *7/21=
  case 4=(r,r,r,r,r)=6/21 * 6/21 * 6/21 * 6/21 * 6/21=
• 8 years agoHelpfull: Yes(2) No(1)
• (6c3*17c2+6c4*17c1+6c5)/21c5
• 8 years agoHelpfull: Yes(1) No(1)
• 6c3 * 15c2+6c4 * 15c1+6c5/21c5
• 8 years agoHelpfull: Yes(1) No(0)
• Total balls =>21
  n(S) => 21C5(as 5 balls are drawn) = 20349
  
  Favourable Cases n(E) => 6C5 + 6C4*8C1 + 6C4*7C1 + 6C3*8C1*7C1 + 6C3*8C2 + 6C3*7C2 = 2331
  
  Probability => n(E) / n(S) = 2331/20349 = 37/327 (ans.)
• 8 years agoHelpfull: Yes(0) No(0)