What is the probability that a leap year has 53 Sundays and 53 Mondays?

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#### Solution

We know that a leap year has 366 days (i.e. 7 \[\times\] 52 + 2) = 52 weeks and 2 extra days.

The sample space for these two extra days is given by

S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}

There are 7 cases.

i.e. *n*(S) = 7

Let E be the event in which the leap year has 53 Sundays and 53 Mondays.

Then E = {(Sunday, Monday) }

i.e. *n*(E) = 1

\[\therefore P\left( E \right) = \frac{n\left( E \right)}{n\left( S \right)} = \frac{1}{7}\]

Hence, the probability in which a leap year has 53 Sundays and 53 Mondays is \[\frac{1}{7} .\]

Concept: Introduction of Event

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