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 atleast three are red!

• 5 ball out of 21 ball can be selected in 21C5 ways.
  at least means favorable events may be (6c3*15c2 + 6c4*15c1 + 6c5)/21c5
• 11 years agoHelpfull: Yes(45) No(20)
• case 1 : 3 red balls : 6x6x6x15x15
  case 2 : 4 red balls : 6x6x6x6x15
  case 3 : 5 red balls : 6x6x6x6x6
  favourable cases = cases ( 1 + 2 + 3) = 30456
  total cases = 21x21x21x21x21 = 21^5
  probability = 376/50421
• 11 years agoHelpfull: Yes(44) No(10)
• @Suraj,chirag : Consider replacement..
  and @Arshal: the method yu hav done is right..but the final answer is 936/50421=312/16807.
• 11 years agoHelpfull: Yes(13) No(2)
• 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
• 10 years agoHelpfull: Yes(11) No(0)
• We have to select 5balls so and atleast 3should be red..
  so more than 3could also be red..
  total ways so selecting 5balls in such a way = 6c3*15c2+ 6c4*15c1+ 6c5
  
  And the total no of ways for selecting 5balls is 21c5
  so the answer will be 37/323
• 11 years agoHelpfull: Yes(6) No(7)
• ((6c3*15c2)+(6c4*15c1)+(6c5))/21c5
• 11 years agoHelpfull: Yes(2) No(5)
• yeah you are right ....i multiplied wrong terms in hurry
• 11 years agoHelpfull: Yes(1) No(2)
• ((6c3*15c2)+(6c4*15c1)+(6c5))/21c5
• 10 years agoHelpfull: Yes(1) No(0)
• pow(5,7)*5
• 11 years agoHelpfull: Yes(0) No(13)
• @Arshal Jain : This is the right answer!!
• 11 years agoHelpfull: Yes(0) No(0)
• ((6c3*15c2)+ (6c4*15c1)+ 6c5)/21c5=2331/20349=259/2261
• 11 years agoHelpfull: Yes(0) No(1)
• As out of 6 red 8 blue and 7 green balls are there we have space = 21
  event is ,
  case 1= 6c5=6( all are red)
  case 2=6c3+8c2=48(3 red and 2 blue)
  case 3= 6c3+7c2=41(3 red and 2 green)
  hence event=6+48+41=95
  so , n(E)/n(s)=95/21
  
  ANSWER= 95/21
• 9 years agoHelpfull: Yes(0) No(3)
• total ways so selecting 5balls in such a way = 6c3*15c2+ 6c4*15c1+ 6c5
  
  
  And the total no of ways for selecting 5balls is 21c5
  so the answer will be 37/323 ""
• 7 years agoHelpfull: Yes(0) No(0)
• 70-54=16 i.e square of 4.
  54-45=9=i.e square of 3
  45-41=4 i.e square of 2
  4,3,2 are following a pttern in decreasing order so
  41-x=1 i.e square of 1.
• 6 years agoHelpfull: Yes(0) No(0)