Question

Probability of encounter 53 sundays in non leap year

Answer 1

$\textbf{Answer}$

We know that,
$\textbf{Every leap year has 366 days}$
&
$\textbf{Every non leap year has 365 days}$

=> Every $\textbf{non leap year}$ has $\textbf{52 weeks and 1 day}$.

This means every non leap year has minimum 52 Sundays.

Remaining 1 day in leap year can be any one of 7 days of a week which can be one of these days -
Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.

So we need to calculate the probability of getting Sunday from the given 7 days.

$\textbf{So probability of 53 Sundays}$ in a $\textbf{Non leap year is 1/7}$.

$\textbf{Required probability is 1/7}$

$\textbf{Hope My Answer Helped}$
$\textbf{Thanks}$

Answer 2

We know that,
\textbf{Every leap year has 366 days}Every leap year has 366 days 
&
\textbf{Every non leap year has 365 days}Every non leap year has 365 days 

=> Every \textbf{non leap year}non leap year has \textbf{52 weeks and 1 day}52 weeks and 1 day .

This means every non leap year has minimum 52 Sundays.

Remaining 1 day in leap year can be any one of 7 days of a week which can be one of these days -
Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.

So we need to calculate the probability of getting Sunday from the given 7 days.

\textbf{So probability of 53 Sundays}So probability of 53 Sundays in a \textbf{Non leap year is 1/7}Non leap year is 1/7 .

\textbf{Required probability is 1/7}Required probability is 1/7