there r three dices one blue and two red.find the probability that the no appered in red dices is more than the blue dice?

• if blue die shows 1, red die can have any of the 5 nos(2, 3, 4, 5, 6) which are greater than 1 hence no of ways = 5C1*5C1
  blue die= 2, red die= any one from (3,4,5,6) hence= 4C1*4C1
  blue die=3, red die= 3C1*3C1
  blue die=4, red die = 2C1*2C1
  blue die=5, red die=1C1*1C1
  total possibilities as per given condition= 25+16+9+4+1=55
  hence probability= 55/216
• 9 years agoHelpfull: Yes(22) No(0)
• We have 3 dices it means total probability is 216. Now the no. on red dices should be more than the blue.
  then the no. on red which are less or equal to blue are..
  (1,1).....(1,5)
  (2,1).....(2,4)
  (3,1).....(3,3)
  (4,1).....(4,2)
  (5,1)
  total 15
  now total outcomes are 216
  therefore favourable is 216-15=201
  hence probability is 201/216=67/72 ans.
• 9 years agoHelpfull: Yes(9) No(9)
• Total possibility - 216
  No. Appeared in red dices more than blue dices- 55
  Probability - 55/216
• 9 years agoHelpfull: Yes(8) No(1)
• considering the case where n(blue)>n(red) we will have 4/18 as 3,4,5 and 6 are the available condition , so the solution should me 1-4/18=14/18
• 9 years agoHelpfull: Yes(4) No(6)
• there are 3 dices.....so total no of cases will be (6*6*6)
  Now if blue dice will have 1,rd dice will have nos more than 1,similarly if blue dice will have 2,red can have 4 nos greater than 2 and in the similar way goes on.So,it will continue upto 5 for blue dice,then red dice will have 6 which is greater than 5.........so no of cases for blue dice having no smaller than red one=(1+2+3+4+5)ie 15.........so (216-15)ie 201
  therefore,prob= 201/216 ie 67/72
• 9 years agoHelpfull: Yes(2) No(1)
• 55/216...is the correct answer...
• 9 years agoHelpfull: Yes(1) No(2)
• While throwing one red dice along one blue dice, we can find 15 ways where red dice numbrs are grater than the blue, hence 15/216. Please correct me if I have mistaken smwhr
• 9 years agoHelpfull: Yes(1) No(5)
• please tell the correct answer.I dont know the probability
• 9 years agoHelpfull: Yes(0) No(0)
• correct ans/////???????
• 9 years agoHelpfull: Yes(0) No(0)
• solve it by bottom-up approach
  first,find the number of ways by which the number on blue dice can be greater than the numbers on red dice
  Blue Red1 Red2 Total_no_of_ways
  1 - - 0
  2 1 1 1
  3 1 1 ......
  3 1 2 4
  3 2 1
  3 2 2
  4 .... ..... 9
  5 ..... ..... 16
  6 ..... ..... 25
  
  total no of ways=(1+4+9+16+25)=55+no_of ways in which all numbers can beequal(=6)=61
  total no ways in which numbers can appear on all 3 dices=6*6*6=216
  probability that numbers on blue dice is greater than or equal to the numbers on red dices=61/216
  Required probabilty=1-61/216=155/216
• 9 years agoHelpfull: Yes(0) No(2)
• if blue die shows 1, red die can have any of the 5 nos(2, 3, 4, 5, 6) which are greater than 1 hence no of ways = 5C1*5C1
  if blue die shows 2, red die can have any of the 4 nos(3, 4, 5, 6) which are greater than 1 hence no of ways = 4C1*4C1
  if blue die shows 3, red die can have any of the 3 nos(4, 5, 6) which are greater than 1 hence no of ways = 3C1*3C1
  if blue die shows 4, red die can have any of the 2 nos(5, 6) which are greater than 1 hence no of ways = 4C1*4C1
  if blue die shows 5, red die can have any of the 1 nos( 6) which are greater than 1 hence no of ways 1C1*1C1
  total possibilities as per given condition= 25+16+9+4+1=55
  hence probability= 55/6^3
• 5 years agoHelpfull: Yes(0) No(0)