Question
What is the probability that an ordinary year has $53$ Sundays$?$
✓
Answer
Given: An ordinary year
To Find: Probability that a non leap year has $53$ Sundays.
Total number of days in an ordinary year is $365$ days
Hence number of weeks in an ordinary year is $\frac{365}{7}=52\text{ weeks and 1 day}$
In an ordinary year we have $52$ complete weeks and $1$ day which can be any day of the week i.e. $\text{SUNDAY, MONDAY, TUESDAY, WEDNESDAY, THURSDAY, FRIDAY and SATURDAY}$
To make $53$ Sundays the additional day should be Sunday
Hence total number of days is $7$
Favorable day i.e. Sunday is $1$
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability that an ordinary year has $53$ Sundays is equal to $\frac{1}{7}$
To Find: Probability that a non leap year has $53$ Sundays.
Total number of days in an ordinary year is $365$ days
Hence number of weeks in an ordinary year is $\frac{365}{7}=52\text{ weeks and 1 day}$
In an ordinary year we have $52$ complete weeks and $1$ day which can be any day of the week i.e. $\text{SUNDAY, MONDAY, TUESDAY, WEDNESDAY, THURSDAY, FRIDAY and SATURDAY}$
To make $53$ Sundays the additional day should be Sunday
Hence total number of days is $7$
Favorable day i.e. Sunday is $1$
We know that $\text{Probability}=\frac{\text{Number of favourable event}}{\text{Total number of event}}$
Hence probability that an ordinary year has $53$ Sundays is equal to $\frac{1}{7}$
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