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Find the probability that a leap year, selected at random, will contain 53 Sundays.

Last updated date: 13th Jun 2024
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Hint: Here we will first use some known facts for e.g. :-
A leap year has 366 days.
A non- leap year with 365 days has 52 weeks which implies it has 52 Sundays.
Hence we will calculate the number of Sundays in a leap year and the find the probability of 53 Sundays by using the formula of probability given by:-
\[{\text{probability}} = \dfrac{{{\text{favourable outcomes}}}}{{{\text{total outcomes}}}}\]

Complete step-by-step answer:
We know that a leap year has 366 days which implies it has 52 weeks and 2 days.
Now we know that 52 weeks have 52 Sundays so we need to find the 53rd Sunday out of 2 days left.
Hence we will make all possible combinations taking 2 days of a week:=
1. {Sunday, Monday}
2. {Monday, Tuesday}
3. {Tuesday, Wednesday}
4. {Wednesday, Thursday}
5. {Thursday, Friday}
6. {Friday, Saturday}
7. {Saturday, Sunday}
Therefore, total outcomes are 7
And since we got two such combinations which have Sunday in them
Hence the favorable outcomes are 2
Therefore now we will calculate the probability by using the following formula:-
\[{\text{probability}} = \dfrac{{{\text{favourable outcomes}}}}{{{\text{total outcomes}}}}\]
Putting in the respective values we get:-
\[{\text{probability}} = \dfrac{2}{7}\]
Therefore the probability that a leap year, selected at random, will contain 53 Sundays is \[\dfrac{2}{7}\]

Note: Students should take a note that the probability of any event is always less than one.
In such questions, we need to make cases or combinations to get the desired answer.