Question

In a non-leap year, the probability of getting $53$ Sundays or $53$ Tuesdays or $53$ Thursdays is $\frac{a}{b},$ where $a$ and $b$ are co-prime. Then the value of $a+b$ is

Answer 1

In a non-leap year, number of days is $365$ i.e., $7\times 52+1.$
$\therefore$ It has $52$ weeks.
Remaining one day can be any day out of $7$ days.
If it's a Sunday, then the probability of $53$ Sundays $=\frac{1}{7}.$
If it's a Tuesday, then the probability of $53$ Tuesdays $=\frac{1}{7}.$
If it's a Thursday, then the probability of $53$ Thursdays $=\frac{1}{7}.$
$\therefore$ Required probability $=\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=\frac{3}{7}$
Hence, $a+b=10$