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For the Question "Find the probability that a leap year has 53 Sundays".

The Solution goes :

For 53 Sundays, we proceed as: $\frac{366}{7} = 52.28$; So we can be sure that there are 52 Sundays, now the only question is of the 1 Sunday. So $7\cdot 52 = 364$. Therefore the remaining 2 days (366-364) decide the probability of being a Sunday. So we consider pairings of 2 consecutive days -

  1. Sunday-Monday
  2. Monday-Tuesday
  3. Tuesday-Wednesday
  4. Wednesday-Thursday
  5. Thursday-Friday
  6. Friday-Saturday
  7. Saturday-Sunday

So out of the 7 pairs, we have 2 pairs that include a Sunday. Therefore the probability of having 53 Sundays in a Leap Year is $\frac{2}{7}$.

Then I was wondering, what if we ask "Probability that a leap year has 52 Sundays", Considering the above solution for 53 Sundays, the answer to the 52 Sundays problem should be 1. (Since $\frac{366}{7} = 52.28$) So we have complete 52 weeks and hence always 52 Sundays, therefore 1.

I found some online forums, where the answer to $52$ Sundays Problem is not 1. Hence my confusion. Please help.

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    $\begingroup$ Does "has 52 Sundays" mean "has at least 52 Sundays" or "has exactly 52 Sundays"? $\endgroup$ – Hagen von Eitzen May 3 '14 at 14:48
  • $\begingroup$ Has exactly 52 Sundays. $\endgroup$ – Abhishek Potnis May 3 '14 at 15:19
  • $\begingroup$ Is the logic to find P(53 Sundays) that I have mentioned correct? $\endgroup$ – Abhishek Potnis May 3 '14 at 15:33
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By the way, the exact probability that a leap year has 53 Sundays is a bit higher than 2/7: The distribution of days of the week repeats exactly every 400 years: January 1st, 2000, was a Saturday, as will be January 1st, 2400. Within these 400 years, there are 97 leap years, 28 of which start on a Saturday or Sunday; so the probability is $28/97\approx 0.28866$ rather than $2/7\approx 0.28571$.

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Probability that leap year has 52 Sundays is 5/7

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if it asked:"What is the Probability that a leap year has exactly 52 Sundays ?", than you can work with the converse probability.

greeting,

calculus

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  • $\begingroup$ So would that be 1 - P(atleast 52 Sundays)? In that case how do I calcualte P(atleast 52 Sundays)? $\endgroup$ – Abhishek Potnis May 3 '14 at 15:29
  • $\begingroup$ No it is $P(X=52)=1-P(X=53)$ -- A year has 52 or 53 sundays. So the probability for at least 52 sundays is one. $\endgroup$ – callculus May 3 '14 at 16:36
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A leap year has 52 weeks + 2 extra days; There are 52 Sundays in 52 weeks; If either of the 2 extra days is Sunday then there will be 53 Sundays in a week; Probability of any one of that extra days is Sunday = Probabilty of 53 Sundays in a week =2/7; Probability That any day of the 2 extra days is not sunday = Probability of only 52 Sundays in a week = 1 - 2/7 = 5/7

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  • $\begingroup$ Use Mathjax please. $\endgroup$ – SchrodingersCat Feb 26 '16 at 13:28

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