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Find the probability that a leap year has 53 Sundays.

Answer
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Hint: Think of the number of complete weeks a leap year has and think about the possible cases for which a leap year can have 53 Sundays.

Complete answer:

Probability $=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ .

To start with the question, let us calculate the number of complete weeks a leap year has.

We know, a week consists of seven days and 366 days form a leap year.
When we divide 366 by 7 we get 52 as the quotient and 2 as the remainder.

Therefore, we can state that a leap year has 52 complete weeks and 2 odd days.
Now, for 53 Sundays, one of the 2 odd days needs to be a Sunday.

All the possible outcomes for the two odd days are:
{(Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), (Sunday, Monday)}.

$\therefore $ The total number of outcomes are = 7.

Among the above outcomes, the favourable outcomes are:
{(Sunday, Monday), (Saturday, Sunday)}
$\therefore $ The number of favourable outcomes = 2.

Using the mathematical definition of probability:
 $\text{Probability}=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
$\Rightarrow \text{Probability}=\dfrac{2}{7}$

Hence, probability that a leap year has 53 Sundays is $\dfrac{2}{7}$

Note: It is preferred that while solving a question related to probability, always cross-check the possibilities, as there is a high chance you might miss some or have included some extra or repeated outcomes.