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. 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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