We have the years from 2001, 2002, 2003,... to 2010. Say, a year is chosen at random from the listed years. What is the probability that the chosen year has 53 Mondays ?
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$\begingroup$ How many of those years have 53 mondays? Count 'em up! $\endgroup$– Mario CarneiroApr 2, 2014 at 14:51
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2$\begingroup$ Well, only 2002 and 2008 had 53 Mondays. That's 2 years out of 10. So if you choose at random one year out of 2001..2010, what's the probability to hit either 2002 or 2008? $\endgroup$– PiwiApr 2, 2014 at 14:53
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2 Answers
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In $400$ years (a Gregorian calendar cycle) there are $365\times 303+366\times 97 =146097$ days which is $\frac{146097}{7}=20871$ weeks and so there are $20871$ Mondays.
Since $20871=329\times 52+71\times 53$, there are $71$ years with $53$ Mondays, and so the probability that a year has $53$ Mondays is $\frac{71}{400}=0.1775$.
Similar calculations over a $400$ year cycle can show that the $13$th of a month is more likely to be a Friday than another other particular day.
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$\begingroup$ The question specifies that the year is between 2001 and 2010. $\endgroup$– MJDApr 2, 2014 at 16:09
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$\begingroup$ Also, Friday 13 and Wednesday 13 are equiprobable. $\endgroup$– MJDApr 2, 2014 at 16:14
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$\begingroup$ @MJD: Friday $13$th: $688$ times in $400$ years. Wednesday $13$th: $687$ times in $400$ years $\endgroup$– HenryApr 2, 2014 at 17:31
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A year with 53 Mondays must start on a Monday, or be a leap year starting on a Sunday, on average a given year having 53 Mondays occurs about once every five or six years (five times in a period of twenty-eight years).
Since 2000 and going up to 2050, the following years have 53 Mondays:
2001 2007 2012 (LY) 2018 2024 (LY) 2029 2035 2040 (LY) 2046
LY = Leap Year
