There are 3 red balls,3 green balls and 3 blue balls are present.what is the probability of atleast two of them are of same colour if 3 balls are drawn?

• Total combinations = 9c3 = 84
  if we calculate the combinations in which three balls are of different colours,
  then combinations will be = 3c1*3c1*3c1 = 27
  so final probability of atleast two of them are of same colour if 3 balls are drawn = 1 - (27/84)
  = 57/84
• 10 years agoHelpfull: Yes(31) No(5)
• 3 balls are drawn two of them are of same color:
  there are three cases (i)red (ii)green (iii)blue
  (i)red means 2 of them are red and any one out of 6 other balls=3c2*6c1
  (ii)green means 2 of them are green and any one out of 6 other balls=3c2*6c1
  (iii)blue means 2 of them are blue and any one out of 6 other balls=3c2*6c1
  initially 3 balls are drawn out of 9 balls=9c3=84
  therefore total probability is (3c2*6c1+3c2*6c1+3c2*6c1)/84=54/84=27/42 is the answer.
  
• 9 years agoHelpfull: Yes(31) No(19)
• atleast 2 of same colour.so there can be 3 of same colour also.
  total probability : 9c3=84
  probabilty that 2 of same colour : 3c2+6c1=18 .there can be 3 such cases so 3(3c2+6c1)= 53
  probablity that 3 of same colour = 3c3=1 and three such cases of 3
  now = 54+3/84 =57/84
  
• 9 years agoHelpfull: Yes(16) No(1)
• 3 balls are drawn two of them are of same color:
  there are three cases (i)red (ii)green (iii)blue
  (i)red means 2 of them are red and any one out of green or blue = (3c2*3c1*3c0)+(3c2*3c0*3c1)
  (ii)green means 2 of them are green and any one outof blue or red =(3c1*3c2*3c0)+(3c0*3c2*3c1)
  (iii)blue means 2 of them are blue and any one out of green or red=(3c0*3c1*3c2)+(3c1*3c0*3c2)
  in the Q it is given atleast two...it means 2 and more than 2 ==>3
  then for all the 3 of same clour is =(3c3*3c0*3c0)+(3c0*3c3*3c0)+(3c0*3c0*3c3)
  initially 3 balls are drawn out of 9 balls=9c3=84
  prbability =((3c2*3c1*3c0)*3)+3)/84 = 57/84
• 9 years agoHelpfull: Yes(7) No(0)
• prob of at least 2 same ball=(1-prob of all diff ball)
  =1-{(3C1*3C1*3C1)/9C3}
  =1-{(3*3*3)/(9!/3!*6!)}
  =1-{(9/84)}
  =75/84
• 9 years agoHelpfull: Yes(6) No(0)
• 3 balls are drawn two of them are of same color:
  there are three cases (i)red (ii)green (iii)blue
  (i)red means 2 of them are red and any one out of green or blue = (3c2*3c1*3c0)+(3c2*3c0*3c1)
  (ii)green means 2 of them are green and any one outof blue or red =(3c1*3c2*3c0)+(3c0*3c2*3c1)
  (iii)blue means 2 of them are blue and any one out of green or red=(3c0*3c1*3c2)+(3c1*3c0*3c2)
  in the Q it is given atleast two...it means 2 and more than 2 ==>3
  then for all the 3 of same clour is =(3c3*3c0*3c0)+(3c0*3c3*3c0)+(3c0*3c0*3c3)
  initially 3 balls are drawn out of 9 balls=9c3=84
  prbability =((3c2*3c1*3c0)*3)+3)/84 = 57/84
• 9 years agoHelpfull: Yes(3) No(0)
• i think ans will b 1/4
• 10 years agoHelpfull: Yes(2) No(4)
• 3 balls are drawn two of them are of same color:
  there are three cases (i)red (ii)green (iii)blue
  (i)red means 2 of them are red and any one out of green or blue = (3c2*3c1*3c0)+(3c2*3c0*3c1)
  (ii)green means 2 of them are green and any one outof blue or red =(3c1*3c2*3c0)+(3c0*3c2*3c1)
  (iii)blue means 2 of them are blue and any one out of green or red=(3c0*3c1*3c2)+(3c1*3c0*3c2)
  in the Q it is given atleast two...it means 2 and more than 2 ==>3
  then for all the 3 of same clour is =(3c3*3c0*3c0)+(3c0*3c3*3c0)+(3c0*3c0*3c3)
  initially 3 balls are drawn out of 9 balls=9c3=84
  prbability =((3c2*3c1*3c0)*6)+3)/84 = 57/84
• 9 years agoHelpfull: Yes(2) No(0)
• Total combinations = 9C3=84
  only one combination is undesirable when three balls are from 3 different colours.
  so
  probability of atleast two of them are of same colour if 3 balls are drawn= 83/84
  
  
• 10 years agoHelpfull: Yes(1) No(9)
• only one combination is undesirable when three balls are from 3 different colours.
  
  3 combinations are there when 3 balls are of same (one) colour.
  6 combinations are there when balls are of 2 colours.
  so total possible combinations =10
  so
  probability of atleast two of them are of same colour if 3 balls are drawn= 9/10
  
  
• 10 years agoHelpfull: Yes(1) No(9)
• N(S)= 9C2
  N(E)= 21
  (SINCE, the combination may be (RRG,GRR,RGR,GGR,GRG,RGG,RRB,BRR.RBR,GGB,BGG,GBG,BBG,BGB,GBB,BBR,BRB,RBB,RRR,BBB,GGG)
  HENCE, P(E)=N(E)/NS)
  i.e. 21/9C3 = 1/4
• 9 years agoHelpfull: Yes(1) No(0)
• total balls 9c3=84
  atleast 3drawn form each balls=(3c3*3c2)+(3c3)=10
  
  
  n(E)/n(s)=84/10=42/5
  
• 10 years agoHelpfull: Yes(0) No(10)
• ans: 53/56
  
  The probability that atleast two of them are of same colour = 1 - probability that 3 balls drawn are different colour
  
  probability that 3 balls drawn are different colour = 3/9 * 3/8 * 3/7
  = 3/56
  
  Therefore
  The probability that atleast two of them are of same colour = 1-3/56
  = 53/56
  
• 10 years agoHelpfull: Yes(0) No(7)
• ans:-27
  
  we have 3g,3b,3r according given condition atleast two of them are of same colour if 3 balls are drawn
  
  3c2*3c1*3c0+3c1*3c2*3c0+3c0*3c1*3c2
  
  9+9+9=27
  
  
  
• 10 years agoHelpfull: Yes(0) No(3)
• answer is 25/28 bcoz total combination is 9c3
  atleast 2 balls having same colour either red nd green is 3c23c1 r green nd red is 3c13c2 like wise green nd blue is 3c23c1 r blue nd green 3c13c2 likewise blue nd red is 3c23c1 r red nd blue is 3c13c2 then add al we get total is 72 r three balls r having same colour either red r green r blue is 3c3+3c3+3c3 =3
  72+3=75/9c3=25/28
  
• 10 years agoHelpfull: Yes(0) No(4)
• 2/3 because total outcomes will be 9 nd favourable will be 6
• 10 years agoHelpfull: Yes(0) No(2)
• atleast two of them are of same colour=3(3c2*9c1/9c3)=42/84
  all of them are of same colour=3(3c3/9c3)=3/84
• 9 years agoHelpfull: Yes(0) No(3)
• 3*3c2/9c2=1/4
• 9 years agoHelpfull: Yes(0) No(3)
• 27/42
  drawn two of them are of same color:
  there are three cases (i)red (ii)green (iii)blue
  (i)red means 2 of them are red and any one out of 6 other balls=3c2*6c1
  (ii)green means 2 of them are green and any one out of 6 other balls=3c2*6c1
  (iii)blue means 2 of them are blue and any one out of 6 other balls=3c2*6c1
  initially 3 balls are drawn out of 9 balls=9c3=84
  
• 9 years agoHelpfull: Yes(0) No(0)
• 6(3c2.3c1)+3.3c3/9c3=57/84=19/28
• 9 years agoHelpfull: Yes(0) No(2)
• Atleast 2 of them must be of same color
  so,we have to find out the probability for 2 or more balls to be of same color.
  so,for 2 or more red balls to be of same color the combinations are ->3c2*7c1.
  similarly,for blue and green balls combinations are the same.
  Also, the total number of balls fetched is 3 out of nine so combination is->9c3
  so, the probability must be {3(3c2*7c1)}/9c3
  which will come out to be 63/84.
  THIS IS THE CORRECT ANSWER
  
• 9 years agoHelpfull: Yes(0) No(2)
• ((3C2*6C1)+3C3)/9C3
• 9 years agoHelpfull: Yes(0) No(0)
• ((3c2*7c1)*3)/9c3=1/4
• 9 years agoHelpfull: Yes(0) No(0)
• ans;9/28
  3 ball drown pro.=84
  3 same color=3times
  atleast 2 same color (ex take RGB == red comes in 2 times RRG RGR GRR RRB RBR BRR so 6 times) 6*3=18
  18+3=21
  ans 21/84===1/4
• 9 years agoHelpfull: Yes(0) No(0)
• correct ans is 5/14 n(s)= 9c3= 84 n(e)= 9*3+3= 30 p(e)=30/84=5/14
• 8 years agoHelpfull: Yes(0) No(0)
• total combination = 9c3 = 84;
  when 2 of the colors are same:
  choose 1 color : 3c1
  choose any 2 balls of that particular color: 3c2
  when 3 colors are same:
  choose any color: 3c1
  total required combination = 3c1*3c2+3c1=12
  probability = 12/84 = 1/7
  
  
  
• 7 years agoHelpfull: Yes(0) No(0)