Question

In a non-leap year the probability of getting $53$ Sundays or $53$ Thursdays is

• A. $\frac{1}{7}$
• B. $\frac{2}{7}$
• C. $\frac{4}{7}$
• D. $\frac{3}{7}$

Answer 1

The correct option is B $\frac{2}{7}$

In anon leap year , there will be $365$ days

$365=7\times 52+1$

Let us assume that the non leap year starts with sunday , then we have $53$ sundays in that year

The next non leap year starts with monday , and has $53$ mondays in that year and so on

So the probability of having $53$ sundays in non leap year is equal to the probability of having $53$ thursdays in non leap year , which is equal to $\frac{1}{7}$

Therefore the required probability is $\frac{1}{7}+\frac{1}{7}=\frac{2}{7}$

So the correct option is $B$