Question

The probability that in a year of the ${22}^{nd}$ century chosen at random there will be $53$ Sundays is

• A. $\frac{5}{28}$
• B. $\frac{7}{28}$
• C. $\frac{11}{28}$
• D. $\frac{13}{28}$

Answer 1

The correct option is A $\frac{5}{28}$

Let $E_1$ be the event it is a non-leap year
and $E_2$ be the event it is a leap year
$P\left(E_1\right)=\frac{3}{4},P\left(E_2\right)=\frac{1}{4}$
As every year will always have $52$ Sundays beacause $52\times 7=364$days
If it is non-leap year number of days will be $365$ days. Hence it's probablity of $53$ Sundays depend upon remaining $1$ day among the set
{Su,Mo,Tu,We,Th,Fr,St} $=P\left(\frac{E}{E_1}\right)=\frac{1}{7}$
If it is leap year number of days will be $366$ days. Hence it probablity of $53$ Sundays depend upon remaining 2 days among the set{SuMo,MoTu,TuWe,WeTh,ThFr,FrSa,SaSu}
$=P\left(\frac{E}{E_2}\right)=\frac{2}{7}$
Using Total Theorem of probablity,
$P\left(E\right)=P\left(E_1\right).P\left(E/E_1\right)+P\left(E_2\right).P\left(E/E_2\right)$
$P\left(E\right)=\frac{3}{4}\times \frac{1}{7}+\frac{1}{4}\times \frac{2}{7}=\frac{5}{28}$