Question

The probability of an ordinary year having $53$ Tuesdays is:

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

Answer 1

The correct option is B $\frac{1}{7}$

$1$ year $=365$ days

A year has $52$ weeks. Hence there will be $52$ Tuesdays for sure.

$52$ weeks $=52\times 7=364$ days

$365-364=2$ days

In an ordinary year, there will be $52$ Tuesdays and $1$ day will be left.

This $1$ day can be:

Sunday

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Of these total $7$ outcomes, the favourable outcomes are $1$.

Hence the probability of getting $53$ Tuesdays in an ordinary year is $=\frac{1}{7}$.