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Find the chance that a non-leap year contains $53$ Saturdays.

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\frac{2}{7}

\frac{1}{7}

\frac{2}{365}

\frac{1}{365}

answer is B.

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Detailed Solution

Given a non-leap year.
Non-leap year has

365

days which is divided into

7 \times 52 = 364 + 1

days.
There will be

52

Fridays, one for each week except when the first day of the year is a Friday.
This means that, we need to calculate the probability of Saturday for 1 day.
So,

n (S) = 7

and

n (E) = 1

,
Probability

P (E) = \frac{n (E)}{n (S)}

Thus, probability is

P (E) = \frac{1}{7}

Therefore, option (2) is correct.

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