Find the probability that a leap year selected at random will have 53 Sundays

• 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
• 9 years agoHelpfull: Yes(19) No(0)
• leap year means 366 days i.e 52 weeks and 2 extra days..
  extra two days may be (Mon,Tue)or(Tue,Wed)or(Wed,Thu)or(Thu,Fri)or(Fri,Sat)or(Sat,Sun)
  i.e n(A)=2 and n(S)=7
  P(n)=n(A)/n(S)=2/7
• 9 years agoHelpfull: Yes(3) No(0)
• give the range of years.
• 9 years agoHelpfull: Yes(1) No(0)
• leap year=52 weeks+2 odd days
  these odd days can be (sat,sun),(sun,mon),(mon,tues),(tues,wed),(wed,thr)(thr,fri)(fri,sat)
  prodability=2/7..
  
• 9 years agoHelpfull: Yes(1) No(0)
• days in leap year 366/7=52,REMAINDER=2;
  so possible days on remaining two days
  (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.
  now total days in a week=7
  probability=2/7 answer.
• 9 years agoHelpfull: Yes(1) No(0)
• ans is 2/7
• 9 years agoHelpfull: Yes(0) No(0)
• ans is 2/7
• 8 years agoHelpfull: Yes(0) No(0)