Question

What is the probability that an ordinary year has \[53\] Sundays? What is the probability that a leap year has \[53\] Tuesdays and \[53\] Mondays?

Answer 1

Hint: In order to find the number of Sundays in an ordinary, firstly we have to consider the number of weeks in a year and then we have check out the number of Sundays we totally get and then consider the favourable outcomes with the possible outcomes. This gives us the required probability. In the same way, in the second case, we will be considering the leap year and check out for the possible outcomes by the number of outcomes and this gives us the next required probability.

Complete step by step answer:
Now let us have a brief regarding the probability. Probability is nothing but how likely an event to occur or how far it is true. The probability of an event always occurs between \[0\] and \[1\]. \[1\] indicates the certainty of an event whereas \[0\]indicates the impossibility of an event. The sum of all the probabilities of an event will always be less than or equal to \[1\].
Now let us find the probability that an ordinary year has \[53\] Sundays.
Ordinary year can also be termed as non-leap year.
Number of days in an ordinary year\[=365\]
Number days in a week\[=7\]
Number of weeks in a year\[=365=7\times 52+1\]
\[\Rightarrow \]\[52\] weeks and one day remains.
The remaining one day can be any day from the week i.e. it can Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.
So the total number of outcomes \[=7\]
From all of the outcomes, the favourable outcome would be any one from the outcomes.
This can be expressed as \[\dfrac{1}{7}\].
\[\therefore \] The probability that an ordinary year has \[53\] Sundays \[=\dfrac{1}{7}\].
Now we will be finding the probability that a leap year has \[53\] Tuesdays and \[53\] Mondays.
Number of days in a leap year \[=366\]
As we know from the above case, the number of weeks would be \[52\] and still \[2\] days will be remaining.
We can say that \[52\] weeks will be having \[52\]Mondays and \[52\] Tuesdays.
So now the remaining two days can be any one from the following combinations:
Sunday, Monday
Monday, Tuesday
Tuesday, Wednesday
Wednesday, Thursday
Thursday, Friday
Friday, Saturday
Saturday, Sunday
We can find that there are seven possible outcomes and any one of the outcomes would be our favourable ones.
\[\therefore \] The probability that a leap year has \[53\] Tuesdays and \[53\] Mondays is \[\dfrac{1}{7}\].

Note: We must always check out for the total possible outcomes so as to calculate the probability for the favourable outcomes. We must also have a note that the sum of the probabilities cannot exceed one.