There are 9 popsicles: 3 orange, 3 cherry, 3 grape. There are 4 children. What is the probability that all 4 children will get the flavor of their choice?

• Consider the 4 children. Each of them has a flavor preference, and so
  there are 3x3x3x3 = 81 possibilities for what the kids want (we assume
  that each kid does in fact want one of the three flavors: orange,
  cherry, or grape).
  
  Of these 81 possibilities, 3 of them mean that one child will be
  unhappy: OOOO, CCCC, and GGGG (i.e., when all four of them want the
  same flavor as the others). For all of the other possibilities (e.g.
  OOCG, OOOC, OOCC,...) there are going to be enough popsicles of the
  required flavors.
  
  Assuming that each of the 81 possibilities is equally likely,
  
  Pr(4 happy kids) = 1 - Pr(at least one unhappy kid)
  = 1 - 3/81
  = 1 - 1/27
  = 26/27
  
  A supply of four of each flavor would have guaranteed four happy kids,
  but a supply of three of each flavor works in almost all (96%) of the
  cases.
• 11 years agoHelpfull: Yes(6) No(0)
• questm iz wrong...IF all the four have same choices..then nly 3 popsciles arre there of the same group..so one of the children will sit empty..which iz not posible.
• 11 years agoHelpfull: Yes(3) No(2)
• Consider the 4 children. Each of them has a flavor preference, and so there are 3x3x3x3 = 81 possibilities for what the kids want Of these 81 Possibilities, 3 of them mean that one child will be unhappy
  Pr(4 happy kids) = 1 - Pr(at least one unhappy kid)
  = 1 - 3/81
  = 1 - 1/27
  = 26/27
  
• 11 years agoHelpfull: Yes(2) No(0)