My name is PREET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?
a) 5 %
b) 10%
c) 20%
d) 25%
e) 12.5%

• 10%
  son can interchange a pair in 10 combination out of which only one keps the name unchanged.
  10 combinations for change are
  PR
  PE
  PE
  PT
  RE
  RE
  RT
  EE
  ET
  ET
  
  so the probability that despite this interchange, the name remains unchanged = 1/10 = 10%
• 12 years agoHelpfull: Yes(62) No(6)
• no of ways in which name remain same is 1 by intrchng e,e.noof ways of intrchng 2 letters is 5C2=10.thre4 probality is 1/10 .so ans is 10%.
  
• 12 years agoHelpfull: Yes(39) No(3)
• preet
  no of sample space in which change can be happen is = {E,E}..........i.e number of possible set = 1
  and total set is = 5 (because 5 letters are there)
  so probability = 1/5
  % probability = 1/5*100
  equals 20
• 12 years agoHelpfull: Yes(10) No(24)
• Ans. 25%
  According to Q "A pair of letters was inter changed. So possible pairs in PREET are
  1.PR
  2.RE
  3.EE
  4.ET
  (Rest combinations did not make a pair in PREET like PE, PT, etc)
  Total 4 out of which only "EE" is the one required
  so probability = 1/4 or 25%.
• 12 years agoHelpfull: Yes(6) No(7)
• ANS : c) 20%
  Here, by interchanging the letters we can construct
  5!/2(divided by 2 because of same letter 'E' is used twice here )
  =60 different names.
  Now his son type his name by interchanging a pair of letter.
  So we can get
  (5-1)C2 (because here E letter is used twice)
  =4C2 = 12 different pair.
  Now our required probability is 12/60 =1/5 or (1/5 * 100%) = 20%
• 12 years agoHelpfull: Yes(4) No(17)
• preet
  no of sample space in which change can be happen is = {E,E}..........i.e number of possible set = 1
  and total set is = 5 (because 5 letters are there)
  so probability = 1/5
  % probability = 1/5*100
  equals 20
• 12 years agoHelpfull: Yes(4) No(9)
• ther are 5 chance of interchanging pair of letters are (P.R),(R,E),(E,E),(E,T),(T,P), but if you interchange (E,E) there is no change in name so probability of same name is 1/5 so ans is 20%
• 12 years agoHelpfull: Yes(3) No(14)
• 10%
  son can interchange a pair in 10 combination out of which only one keps the name unchanged and that is EE
  10 combinations for change are
  PR
  PE
  PE
  PT
  RE
  RE
  RT
  EE
  ET
  ET
  
  so the probability that despite this interchange, the name remains unchanged = 1/10 = 10%
  
• 10 years agoHelpfull: Yes(1) No(0)