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What is the probability that a non-leap year has 53 Sundays?

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Hint- In this question we need to find the probability that a non-leap year has 53 days, so a leap year is one which has 366 days instead of a normal year which has only 365 days. So use this concept along with the concept of probability to obtain the answer.

Now we have to find the probability that a non-leap year has 53 Sundays.
So,
As we know that a non-leap year has 365 days also there are 7 days in a week.
Since a non-leap year has 365 days and a week has in total 7 days, so let's compute the number of weeks in a non-leap year, hence it will be 365 divided by 7
So when we divide 365 by 7 we get 52 as quotient and 1 as remainder.
So, there are 52 weeks and 1 extra day.
So, there will be only one such day out of 7 which will occur for 53 times in a non-leap year.
We have to find the probability that the day which occurs 53 times is Sunday.
Let A be an event of getting 53 Sundays in a non-leap year.
So n(A)=1………………………. (1)
Since a week has only 7 days, and we are considering one Sunday out of these 7 days hence the total number of outcomes will be 7.
So n(S) = 7………………………. (2)
As we know that probability is defined as the number of favourable outcomes divided by the number of Total Outcomes, that is
${\text{Probability = }}\dfrac{{n\left( A \right)}}{{n\left( S \right)}}$
So putting values in the above formula using equation (1) and equation (2) we get $\dfrac{1}{7}$.
Thus the probability that a non-leap year has 53 Sundays is $\dfrac{1}{7}$.

Note- To solve these types of questions we must remember that a non-leap year has 365 days. Then when we compute the remainder of 365 divided by 7 we get one. So, we get to know that only one day will repeat for 53 times. Then according to the question we can find the favourable outcomes out of the total outcomes. Using which we compute the probability of the given event.
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