On: What is the Probability of 53 Sundays in an Ordinary Year?

  • Thread starter Thread starter suganya
  • Start date Start date
AI Thread Summary
The probability of an ordinary year containing 53 Sundays is determined by the starting day of the year. An ordinary year has 365 days, which equates to 52 weeks and 1 extra day. If the year starts on a Sunday, that extra day will also be a Sunday, resulting in 53 Sundays. Since there are seven possible starting days of the week, the chance of a year starting on a Sunday is 1 in 7. Therefore, the probability of having 53 Sundays in a non-leap year is approximately 14.29%.
suganya
Messages
4
Reaction score
0
Hi,

What is the chance that an ordinary year selected at random contains53 Sundays.

How to work it.

Regards,
Suganya

EasyCalculation
ToFocus
 
Physics news on Phys.org
If by ordinary year you mean non-leap year, then under what circumstances will a year have 53 Sundays?
 
What if the year starts on a Sunday? What if it starts on a Monday? How often does it start on each day in a given calendar (just assume 1/7 for a first estimate)?
 
Well, a general year here is 365 days. Dividing it into weeks (by 7) would leave a remainder of 1. Where there would be 52 weeks and 1 day. We can assume that those 52 weeks would mean 52 Sundays (now working with an imaginary 364 day-year, just because it doesn't leave the remainder of 1). To have the 53 Sundays is when the remainder of 1 comes in. Suppose you had our imaginary year of 364 days and let's say it began on a Monday, thus neatly ending on a Sunday. Should we add the remainder of 1 at the end, there will be an extra Monday (Monday now = 53). To be more relevant to your case, it would be easier to start this with a 364 day-year that ended with a Saturday, and thus began with a Sunday. (Imagine moving a 364 day interval left, which is set on a continuous background of Monday-Sunday cycles). So now we add the remainder, to get 365 days, and we have a Sunday at the end (Sunday now = 53). Should we move this year left, it will be a year that starts with a Saturday and ends with a Saturday (original Sunday still contained, but the one at the end is cut off, now Sunday = 52). We could have started this by adding that remainder to the beginning of the year, and it wouldn't have made a difference. Point is, a 365 day year begins with the same day it ends. That last day makes a day of the week = 53, compared to the other 6/7 days which are at 52. Thus the question can be reduced to - "What are the chances that a year starts with a Sunday?".(assuming you mean a non-leap year). There are 7 choices, hence 1/7. Hope I'm right, as I'm half asleep.
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...
Back
Top