1. ## Probability

Hi,

I am new to this forum. Pls help with my problem.

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

How to work it.

Regards,
Suganya
Easy Calculation
Internet Statistic Report

2. Hello, suganya!

What is the chance that an ordinary year selected at random contains 53 Sundays?

An "ordinary" year has 365 days.

Note that: .$\displaystyle 365 \:=\:52(7) + 1$
. . That is, a year has 52 full weeks plus one day.

This means that a year begins on a certain weekday
. . and also ends on the same weekday.

Hence, a year starting on Sunday will have 53 Sundays.

Assuming the starting day for a year is evenly distributed,
. . the probability of beginning on a Sunday is $\displaystyle \frac{1}{7}$

3. Originally Posted by Soroban
Hello, suganya!

An "ordinary" year has 365 days.

Note that: .$\displaystyle 365 \:=\:52(7) + 1$
. . That is, a year has 52 full weeks plus one day.

This means that a year begins on a certain weekday
. . and also ends on the same weekday.

Hence, a year starting on Sunday will have 53 Sundays.

Assuming the starting day for a year is evenly distributed,
. . the probability of beginning on a Sunday is $\displaystyle \frac{1}{7}$
But does the existance of leap years alter this?

RonL