In a non-leap year, the probability of gettin | vedclass

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(C) non-leap year has $365$ days, which is equal to $52$ complete weeks and $1$ extra day.This extra day can be any of the $7$ days of the week (Sunda

Vedclass · Accepted Answer

$\frac{3}{7}$

Vedclass · Answer

(C) non-leap year has $365$ days, which is equal to $52$ complete weeks and $1$ extra day.This extra day can be any of the $7$ days of the week (Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday).Let $E_1$ be the event of getting $53$ Sundays, $E_2$ be the event of getting $53$ Tuesdays, and $E_3$ be the event of getting $53$ Thursdays.Since these events are mutually exclusive, the probability of getting $53$ Sundays $OR$ $53$ Tuesdays $OR$ $53$ Thursdays is $P(E_1 \cup E_2 \cup E_3) = P(E_1) + P(E_2) + P(E_3)$.$P(E_1) = \frac{1}{7}$, $P(E_2) = \frac{1}{7}$, and $P(E_3) = \frac{1}{7}$.Therefore, the required probability is $\frac{1}{7} + \frac{1}{7} + \frac{1}{7} = \frac{3}{7}$.

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

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