A father purchases dress for his three daughter. The dresses are of same color but of different size .the dress is kept in dark room .What is the probability that all the three will not choose their own dress...




a) 2/3 b) 1/3 c) 1/6 d) 1/9

• a dosesnot choose his dress 2/3
  b dosesnot choose his dress 1/2
  c dosesnot choose his dress 1
  total 1/3
• 12 years agoHelpfull: Yes(109) No(17)
• let the right sequence be A B C
  NOW there are 6 possibilities for the 3 choosing their dress
  A B C
  A C B
  B A C
  B C A
  C A B
  C B A
  Here in the question its asked the probability of all the 3 not to choose there own dress that means 1 can choose her own dress
  total unfavourable outcomes are
  B C A AND C A B
  THEREFORE
  PROBABILITY = UNFAVOURABLE OUTCOMES/TOTAL POSSIBILITIES
  ie 2/6 = 1/3
• 12 years agoHelpfull: Yes(60) No(3)
• ans wil b 2/3..
  probabilty always 1..
  so getting right dresses is 1 out of three 1/3
  so nt getting right will b 1-1/3= 2/3..
  
• 12 years agoHelpfull: Yes(47) No(28)
• 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
• 12 years agoHelpfull: Yes(29) No(4)
• let the three b A B C
  the possibilities are... :
  1-- > all three take the right dress or
  2-- > A takes B's...B takes c's and C takes A's dress or
  3-- > A takes c's...B takes A's..and C takes B's dress...
  so the probabilty that all three will not choose their own dress=2/3
  hence ans is a) 2/3
  
• 12 years agoHelpfull: Yes(6) No(5)
• there are only 2 cases which are favourable hence the answer is 2/6=1/3
• 12 years agoHelpfull: Yes(5) No(3)
• Ans. (b) 1/3
  
  See there are only 3 possibilities
  1. All 3 get correct Dress
  2. Any 1 get correct Dress
  3. No one gets correct dress.
  So Probability will be 1/3 ( As for 3rd option only we will not take 2nd option
  as Not all three have wrong dress in 2nd option).
  
  We have to also ignore - 2 girls get correct dress as if so then 3rd will
  automatically correct. (i.e.. 0, 1, 3) Then prob of 0 = 1/3.
• 12 years agoHelpfull: Yes(5) No(0)
• 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
  
• 9 years agoHelpfull: Yes(5) No(0)
• there's only 3 ways in which the three girls can select their dress.....and one of the way being the correct one...so there's 2 ways in which all the 3 girls will not choose their own dreeses....so the answer is 2/3.
  
• 12 years agoHelpfull: Yes(3) No(9)
• its 1/3 or 2/3 and how?
• 12 years agoHelpfull: Yes(2) No(1)
• plz explain clearly whether it's answer is 1/3 or 2/3 an how??
• 12 years agoHelpfull: Yes(1) No(0)
• b)2/3 because total permutations can be made is 3p3= 6, out of those only 2 are the chances where all the three will not get their dresses
• 12 years agoHelpfull: Yes(1) No(3)
• I don't think 1/3 is the right answer.
  Reason: As per the question, at most 2 girls can choose their own dress, and not all three. There is only one way out of 6 ways where all the three girls will select their right dress. So, the required probability is: 1-1/6=5/6.
  
• 11 years agoHelpfull: Yes(1) No(3)
• ANSWER WOULD BE 1/3
• 12 years agoHelpfull: Yes(0) No(6)
• answer is 1/3 or 2/3 ?
  
• 12 years agoHelpfull: Yes(0) No(0)
• I think the ans is 1/6
  
  because there is only one correct combination.....i.e A B C
  in
  A B C
  A C B
  B A C
  B C A
  C A B
  C B A
  
  so otal posibilities are 6 and faverable event is one....there4 ans should be 1/6
• 12 years agoHelpfull: Yes(0) No(3)
• probability of choose=1/3
  probability that all will not choose=1-1/3=2/3
• 12 years agoHelpfull: Yes(0) No(1)
• total event= 3!=6
  no of probability= 1/6
• 11 years agoHelpfull: Yes(0) No(0)
• 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
• 11 years agoHelpfull: Yes(0) No(0)
• answer is 1/3
• 11 years agoHelpfull: Yes(0) No(0)
• the ans is b
  
• 11 years agoHelpfull: Yes(0) No(0)
• total event= 3!=6
  no of probability= 1/6
• 11 years agoHelpfull: Yes(0) No(0)
• each girl have a chance to select only 2 dress as non of them select their own dress. so total no of selection is 6 so ans is 2/6=1/3
• 9 years agoHelpfull: Yes(0) No(0)
• 1/3 is right ans.
  total no. of ways to select the dress is 3p3=6
  now let three girls are A B and C and thier respective dresses are a b and c.
  as in question none of the girl select her own dress therefor only two possibilities
  A>b B>c C>a
  A>c B>a C>b
  hence probability will be 2/6=1/3
  
  non
• 8 years agoHelpfull: Yes(0) No(0)