What is the probability of 53 sundays in a leap year ?

• A leap year has 366 days, therefore 52 weeks i.e. 52 Sunday and 2 days.
  
  The remaining 2 days may be any of the following :
  
  (i) Sunday and Monday
  
  (ii) Monday and Tuesday
  
  (iii) Tuesday and Wednesday
  
  (iv) Wednesday and Thursday
  
  (v) Thursday and Friday
  
  (vi) Friday and Saturday
  
  (vii) Saturday and Sunday
  
  For having 53 Sundays in a year, one of the remaining 2 days must be a Sunday.
  
  n(S) = 7
  
  n(E) = 2
  
  P(E) = n(E) / n(S) = 2 / 7
• 10 years agoHelpfull: Yes(24) No(0)
• A leap contain 366 days means 52 weeks and 2 odd days.
  Those two days may be Saturday and Sunday or Sunday and Monday
  A week contain 7 days.
  So the probability is:2/7
• 10 years agoHelpfull: Yes(3) No(0)
• A non leap year contains 364 days i.e. 52 weeks..
  A leap year contains 366 days i.e. 52 weeks + 2 days
  These 2 days may have combinations like- (Sunday, Monday) ;(Mon, Tue); (Tue, Wed); (Wed, Thurs); (Thurs, Fri); (Fri, Sat); (Sat, Sun)..
  So there are Seven possibilities of two days in a leap year..
  Therefore probability of getting 53 sundays would be- 2/7
  
  Answer is- 2/7
• 10 years agoHelpfull: Yes(1) No(0)
• 2/7 is the probability.
  
• 10 years agoHelpfull: Yes(1) No(0)
• leap year have 366 days. in this days 52 weeks + 2 days
  2 days can{ M T, T W , W TH , TH F ,F S, S SU , SU M}
  
• 10 years agoHelpfull: Yes(0) No(0)
• As In a leap year there are 366 days with 52 Sundays in a normal year.so the last day of the leap year combining with other days of the week will be SM,MT,TW,WT,TF,FS,SS where all the letters used are the days of a week..so
  Total combination= 7 which is the sample space
  And no.of Sundays has 2 combination SM,SS so answer is 2/7
  
• 10 years agoHelpfull: Yes(0) No(0)
• 2/7
  Sunday & Monday
  Saturday & Sunday
  
• 10 years agoHelpfull: Yes(0) No(0)
• 366 mod 7 = 2 odd days
  2/7 = .285
  
• 10 years agoHelpfull: Yes(0) No(0)
• 2/7 is the ans..
• 9 years agoHelpfull: Yes(0) No(0)