What is the probability that an ordinary year has $53$ Sundays?

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Solution

An ordinary year has $365$ days

$365 \Rightarrow 52 \times 7 = 364 +$ $1$ day is remaining

This One-day can be Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday

Probability of having the Sunday is $\Rightarrow \frac{1}{7}$

Hence, probability is an ordinary year having the Sunday is $\Rightarrow \frac{1}{7}$

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