find the probability that a leap year choosen at random will ave 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(15) No(0)
• p(E)=n(E)/n(7)
  p(E)=2/7
  
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• 366/7=1
  1 day is to be selected out of 7 days
  so ans = 1/7
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• p(E)=n(E)/n(7)
  p(E)=2/7
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• answer=2/7
  
  a leap year has 366 days.52 complete weeks will have 364 days.
  365th day can be sunday or 366th day can be sunday.
  therfore probability(1/7+1/7)=2/7
• 9 years agoHelpfull: Yes(0) No(0)
• Non leap year= 52weeks+1 odd day
  Leap year= 52weeks+2 odd days
  Now out of these 2 odd days in a leap year, sunday could fall on any of the 2 days.
  And, total no. of days in a week= 7days
  Therefore, Probability= 2/7
• 9 years agoHelpfull: Yes(0) No(0)
• 366%7=2
  so ans is 2/7
  
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