Question

A probability that in a year of the ${{22}^{nd}}$ century chosen at random there will be 53 Sundays is
(1) $\dfrac{3}{28}$
(2) $\dfrac{2}{28}$
(3) $\dfrac{7}{28}$
(4) $\dfrac{5}{28}$

Answer 1

We are given a year in the ${{22}^{nd}}$ century and we are asked to find the probability of getting 53 Sundays. We will first find the probability of getting a leap year and a non – leap year. A leap occurs every four years. And a leap year has 52 weeks and 2 days whereas a non – leap year has 52 weeks and 1 day. We will take up separately for Sundays in the additional days of leap year and non – leap year and then we will add them up. Hence, we will have the required probability.

Complete step by step solution:
According to the given question, we are asked to choose a year in the ${{22}^{nd}}$ century and we have to find the probability of having 53 Sundays.
Firstly, we will find the probability of getting a leap year or a non – leap year.
A leap year has 366 day and comes after every four years and that means in 100 years, these are approximately 25 leap years. So, the probability will be,
Probability of getting a leap year in 100 years = $\dfrac{25}{100}=\dfrac{1}{4}$
And the probability of getting a non – leap year = $1-\dfrac{1}{4}=\dfrac{3}{4}$
Also, we know that in a leap year, there are 52 weeks and 2 days. And in case of a non – leap year, there are 52 weeks and 1 day.
For a leap year, we have two additional days and the possibilities that we can have is,
(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
So, the probability of getting 53 Sundays in a leap year = $\dfrac{2}{7}$
Similarly, the probability of getting 53 Sundays in a non – leap year = $\dfrac{1}{7}$
Now, we will take the values of leap years and non – leap years separately and add them up and we get the overall probability as,
$= \left( \dfrac{1}{4} \right)\left( \dfrac{2}{7} \right)+\left( \dfrac{3}{4} \right)\left( \dfrac{1}{7} \right)$
Multiplying the numerators together and the denominators together, we have,
$-= \dfrac{2}{28}+\dfrac{3}{28}$
$= \dfrac{5}{28}$
Therefore, the correct option is (4) $\dfrac{5}{28}$.

Note: Leap year and a non – leap year should be known clearly else the mismatching will result in a wrong answer. While calculating the overall probability, make sure that the values pertaining to leap years are kept together and multiplied correctly as well as the values pertaining to a non – leap year should also be kept together and multiplied. It is advisable to carry out the procedure in different steps separately to avoid mixing up.