find the probability of the leap year choosen randomly will have 53 sundays?

• since leap has 52 week and 2 odd days.
  so 2 odd days may be as follows:-
  sun mon
  mon tue
  tue wed
  wed thr
  thr fri
  fri sat
  sat sun
  so total sample is 7
  and 2 favorable sample that is sun mon and sat sun so
  probality of a leap year choosen randomly wil have 53 sundays is 2/7
  Ans :-2/7
• 9 years agoHelpfull: Yes(2) 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
  Therefore, probability that a leap year has 53 Sundays is 2/7.
  (Note that this is true for any day of the week, not just Sunday)
• 9 years agoHelpfull: Yes(2) No(0)
• 2/7 because Sunday is repeated in twice
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• 2/7 is ans
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• 2/7....................................................................................................................................................................
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• Pls Explain
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