A father purchased dress for his 3 daughters. The dresses are of same color but diff size and they are kept in dark room. what is probability that all the 3 will not choose their own dress?

• the arrangement will follow this way =>first girl will choose wrong dress AND second girl will wrong dress AND the third girl will choose wrong dress.
  since there are three dresses so .
  =>first girl will choose wrong = 1-(1/3) = 2/3
  now two dress are left
  =>second girl will choose wrong out of two dress left => 1-(1/2) =1/2
  now third girl will chose in only one way=1
  so required probability= 2/3*1/2*1
  =1/3 (ans)
• 9 years agoHelpfull: Yes(19) No(0)
• total number of dresses =3
  probabilty of choosing the right dress=(1/3)(1/3)(1/3)= 1/27
  not choosing the right dress = 1-(1/27)= 26/27
• 9 years agoHelpfull: Yes(7) No(7)
• This is a case of de-arrangements = Dn=n!(1/2!−1/3!+1/4!−....)
  So number of ways that none of them chooses their own dress = D3=3!(1/2!−1/3!)=2
  So probability = 2/3!=1/3
  
  :)
• 9 years agoHelpfull: Yes(6) No(1)
• ans: 2/6 = 1/3
  as their are total 6 possibilities of selecting dress (abc, acb, bac, bca, cab, cba) for (g1g2g3) daughter
  suppose a dress for g1 b dress for g2 and c dress for g3 are correct choices
  then out of 6 possibilities only (bca and cab) 4 t and 5 th outcomes are for all the 3 will not choose their own dress.
• 9 years agoHelpfull: Yes(1) No(0)
• 8/27(2/3*2/3*2/3)
• 9 years agoHelpfull: Yes(0) No(1)
• $ $ $ -> dress
  ! ! ! -> girl 1st girl pick either 2nd or 3rd
  2nd girl pick either 1st or 3rd
  3nd girl pick either 1st or 2nd each girl can pick 2 dress ,then 3 girls pick 6 dress. so no.of way 2/6=1/3
• 9 years agoHelpfull: Yes(0) No(0)
• This is a case of de-arrangements = Dn=n!(12!−13!+14!−....)
  So number of ways that none of them chooses their own dress = D3=3!(12!−13!)=2
  So probability = 23!=13
  
• 9 years agoHelpfull: Yes(0) No(0)