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 there own dress?

• probability is 1/3 because
  let 1st girl come and she choose wrong dress so probability of that girl to choose wrong dress out of 3 is =2/3.
  now 2nd girl come nd she choose wrong dress so probability of that girl to choose wrong dress out of 2 is 1/2.
  now for 3rd girl probability is 1 to choose wrong dress.
  so probability tht all the 3 wil not choose der own dress is =2/3*1/2*1=1/3.
  ans is 1/3.
• 10 years agoHelpfull: Yes(153) No(6)
• p(not choose own dress)=1-p(choosing own dress)
  
  p(choose own dress)=(1/3)(1/2)(1)
  =1/6
  
  ans: p(not choose own dress=1-(1/6)
  =5/6
• 10 years agoHelpfull: Yes(31) No(37)
• Total selections=3!=6
  
  Proper Selections:
  A B C
  A C B
  C B A
  B A C (ie. total 4)
  
  Improper selection: B C A , C A B (ie. total 2)
  Probability for improper selection= 2/6 = 1/3
  
• 10 years agoHelpfull: Yes(25) No(4)
• ( 2/3 * 1/2 * 1 )= 1/3 this is the probability that they will not choose their own dress
• 10 years agoHelpfull: Yes(13) No(5)
• (2/3)*(1/2)*1=1/3
• 10 years agoHelpfull: Yes(4) No(1)
• correct answer is 1/3
  
  Method 1)
  3 daughters (correct pairs daughter no. 1-a 2-b 3-c)
  probability that daughter 1 will choose correct dress (i.e a) is 1/3
  probability that daughter 1 will not choose a is 1- 1/3 = 2/3.
  remaining dresses 2 out of which only 1 pair is correct (as first daughter picked someone else's dress)
  so if either 2 or 3 pick incorrect dress with probability of 1/2 last pair is guaranteed wrong
  total probability = 2/3 * 1/2 * 1 = 1/3
  
  
  
  method 2)
  let A`s dress named A, B dress named B, C`s dress named C.
  Then
  Total Combinations are 3!=6
  
  A B C
  A C B
  C B A
  B A C
  B C A
  C A B
  
  Now, We have to find a condition where
  A does not choose her dress,
  B does not choose her dress and
  C does not choose her dress.
  
  In first four condition,
  either A or B or C choose either of its right choice
  like in ACB, A choose right choice... and so on
  but in last two cond.
  B C A
  C A B
  None of them chooses right choice
  So fav outcome=2
  and total outcome=3!=6
  prob-2/6=1/3
• 8 years agoHelpfull: Yes(3) No(0)
• for 1st one= 1/3 second one 1/2 third one=1 total=1/6
• 10 years agoHelpfull: Yes(2) No(3)
• ans is 5/6
• 10 years agoHelpfull: Yes(2) No(5)
• ans is 5/6
• 10 years agoHelpfull: Yes(2) No(5)
• vinay kumar's ans is correct and explanation too
  ans is 1/3
• 10 years agoHelpfull: Yes(2) No(0)
• Formula for case of de-arrangements = n!(1/2! - 1/3! - ... - 1/n!)
  hence, no. of ways dt none of them chooses there own dress = 3!(1/2!-1/3!)=2
  
  P=2/3!=2/6=1/3
• 10 years agoHelpfull: Yes(2) No(1)
• 5/6
  as 1/6 is the probabilty of choosing thier specific dress ,so 1-1/6 is d ans
• 10 years agoHelpfull: Yes(1) No(1)
• 1/9+1/9+1/9=1/3
• 10 years agoHelpfull: Yes(0) No(1)