there are 5 envelope and 5 letters placed at random. what is the probability that no single letter is placed in the right envelope

• probability that no single letter is placed in the right envelop is = 1- (placing all the letters in the right envelop)
  i.e 5 letters will go in their right envelops in only 1 way, and total cases is 5!
  so, 1-(1/5!) = 119/120
• 9 years agoHelpfull: Yes(14) No(1)
• If we let A be the event that letter A is in ist correct envelope and
  similarly B is the event that letter B is in its correct envelope,
  then
  
  P(A) = 1/5 and
  
  P(A and B) = 1/5 * 1/4
  
  Now use the inclusion-exclusion principle to get probability that A or
  B or C .... or E are correctly placed.
  
  P(A or B or C .... or E) = P(A) + P(B) + P(C) + P(D) + P(E)
  - P(A and B) - P(B and C) -....
  + P(A and B and C) + P(B and C and D) + ....
  - P(A and B and C and D) - P(...) -......
  + P(A and B and C and D and E)
  
  = 5*(1/5)
  - 5C2*(1/5)(1/4)
  + 5C3*(1/5)(1/4)(1/3)
  - 5C4*(1/5)(1/4)(1/3)(1/2)
  + (1/5)(1/4)(1/3)(1/2)(1/1)
  
  5*4 1 5*4*3 1 5*4*3*2 1 1
  = 1 - --- * --- + ----- * ----- - --------- * ------- + ---------
  1*2 5*4 1*2*3 5*4*3 1*2*3*4 5*4*3*2 5*4*3*2*1
  
  = 1 - 1/2! + 1/3! - 1/4! + 1/5! = 19/30
• 9 years agoHelpfull: Yes(9) No(10)
• p of placing them in right envelopes is 1/5! hence for not placing them is 1-1/5!=119/120
• 9 years agoHelpfull: Yes(4) No(2)
• case of total disagreement so formula is n!(1/2!-1/3!+1/4!........+1/n!)
  so 5!(1/2!-1/3!+1/4!-1/5!)=44
  so p=44/5!=44/120=11/30.
  
• 9 years agoHelpfull: Yes(2) No(0)
• only 15 cases(1+2+3+4+5) are there in which each letter will go into correct envelope, while total 5^5 cases are there to place the letters into envelope....
  ans is : 5^5 - 15.
• 9 years agoHelpfull: Yes(0) No(3)
• anyone pls explain this prob clearly.
• 9 years agoHelpfull: Yes(0) No(1)
• here it is said no single letter should be at correct place
  so 1stly chances of getting 1 letter at correct place=5
  2ndly chances of getting 2 letters at correct place=5c2=10
  3rdly chances of getting 3 letters at correct place=5c3=10
  lastly chances of getting 5 letters at correct place=1
  note that 4 letters cant be at rt place bcoz it will ultimately lead all letters at correct place
  sum=5+10+10+1=26
  chance of no letter at correct place=120-26=94
  p=94/120
• 9 years agoHelpfull: Yes(0) No(1)
• question is all of them are wrong not only one of them
• 9 years agoHelpfull: Yes(0) No(0)