Find the probability that a leap year chosen at random has 53 sundays?

• Ans:2/7
  A leap year has 366 days
  52 weeks and 2 more days
  The last 2 days can be Following from
  (sunday,monday)
  (monday,tuesday)....(saturday,sunday)
  There are 7 combination
  2 scenarious(days) out of 7 have 53rd sundays
• 10 years agoHelpfull: Yes(40) No(3)
• Sol.: 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(12) No(1)
• 2/7
  because leap year has 2 odd days so probability of a sunday is 2/7
• 10 years agoHelpfull: Yes(2) No(0)
• 366=7*52+2
  the possibility is Sunday&Monday, Mon&Tue, Tue&Wed, Wed&Thu, Thu&Fri, Fri&Sat, Sat&Sun SO THE PROBABILITY IS 2/7
• 10 years agoHelpfull: Yes(2) No(1)
• ans is 2/7
  
• 10 years agoHelpfull: Yes(1) No(0)
• in leap year there are 366days.
  52 weeks means 364 days.
  in one year there are 52 weeks so 52weeks+2days
  2odd days sat+sun,sun+mon
  so 2/7
• 10 years agoHelpfull: Yes(1) No(1)
• ans is 2/7.
  366=7*52+2.
  
• 10 years agoHelpfull: Yes(0) No(0)
• there are 52 weeks + 2 days in a leap year(366=52*7 + 2).ie 52 sundays.
  these extra 2 days can be sun+mon , mon+tue,......sat+sun.so to have 53 sundays one of these consecutive must be a sunday. there are 2 such possibilities out of 7 possibilities. hence reqd probability = 2/7
• 10 years agoHelpfull: Yes(0) No(0)
• 2/7 is the ans
  beacuse 52*7=364 the 2 days are left
  so there are 2 chances of occuring of sunday.
• 10 years agoHelpfull: Yes(0) No(0)
• 2 day
  leap yr has 366 days
  so52*7=364
  2 days can be in pair of sat-sun and sun-mon pair
• 10 years agoHelpfull: Yes(0) No(0)