the probability that there are 53 sundays in a randomly chosen leap year

• assuming 52 sundays in a year are their ... it makes
  52 X 7=364 days
  
  now the remaining 2 days one of the day might be sunday
  
  the probability will be 2/7= 0.28
• 9 years agoHelpfull: Yes(2) No(0)
• 2/7 ....bcoz in a leap year their is 2 odd days....so it may be (sun mon) or (mon tues) and so on...so out of 7 days fav cases are 2...2/7 is ans
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• 2/7 52*7=364 2 odd days rem they can be any of 7 so 2 out of 7 ie 2/7
• 8 years agoHelpfull: Yes(0) No(0)
• Our sample space is S : {Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday,..., Sunday-Monday}
  Number of elements in S = n(S) = 7
  What we want is a set A (say) that comprises of the elements Saturday-Sunday and Sunday-Monday i.e. A : {Saturday-Sunday, Sunday-Monday}
  Number of elements in set A = n(A) = 2
  By definition, probability of occurrence of A = n(A)/n(S) = 2/7
  
• 8 years agoHelpfull: Yes(0) No(0)
• 2/7
  in leap year 366 days.in that 52 sundays are there and two days will remain
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• 366 days in leap year in which 52w and 2d[(sun,mon) (mon,tue) (tue,thu,) (wed,thu) (thu,fri) (fri,sat)and(sat,sun)].so answer is 2/7
• 8 years agoHelpfull: Yes(0) No(0)