MathematicsWhat is the probability that there are 53 Wednesdays in a leap year?

What is the probability that there are 53 Wednesdays in a leap year?


  1. A
    27
  2. B
    37 
  3. C
    57   
  4. D
    17  

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    Solution:

    A leap year is given.
    We know that the probability is given as the ratio of the number of favorable outcomes with the total number of possible outcomes.
    P(E)=Number of favourable outcomes n(E)Total possible outcomes n(S)
    We know that a leap year has 52 weeks and 2 extra days.
    Therefore 52 Wednesdays are present in 52 weeks.
    Total outcomes for the 2 extra days are: Sun, Mon Mon, Tue Tue, Wed , Wed, Thu Thu, Fri Fri, Sat , Sat, Sun .    The total possible outcome, n(S)=7  Let E be the event that represents there are 53 Wednesdays in the year.
    A number of favourable outcomes for 53rd Wednesday are: E={(Wed,Thu),(Tue,Wed)}   The number of elements n(E) that are contained in E, n(E)=2 The probability of the event E is,
    P(E)=Number of favourable outcomes n(E)Total number of outcomes n(S)
    P(E)=27
    Hence, the probability that there are 53 Wednesdays in the year are 27.
    Therefore, option 1 is correct.
     
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