The probability that a non-leap year selected | vedclass

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Question

(C) non-leap year has $365$ days,which is $52$ weeks and $1$ extra day. The extra day can be any one of the $7$ days of the week: {Monday,Tuesday,Wedn

Vedclass · Accepted Answer

$\frac{2}{7}$

Vedclass · Answer

(C) non-leap year has $365$ days,which is $52$ weeks and $1$ extra day. The extra day can be any one of the $7$ days of the week: {Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday}. Let $A$ be the event that the year has $53$ Saturdays and $B$ be the event that the year has $53$ Sundays. Since the year has $52$ weeks,it already contains $52$ Saturdays and $52$ Sundays. For the year to have $53$ Saturdays,the extra day must be a Saturday. $P(A) = \frac{1}{7}$. For the year to have $53$ Sundays,the extra day must be a Sunday. $P(B) = \frac{1}{7}$. Since the extra day cannot be both a Saturday and a Sunday simultaneously,$A$ and $B$ are mutually exclusive events. The probability that the year has $53$ Saturdays or $53$ Sundays is $P(A \cup B) = P(A) + P(B) = \frac{1}{7} + \frac{1}{7} = \frac{2}{7}$. Note: The question asks for $52$ Saturdays $OR$ $53$ Sundays. Since every non-leap year has $52$ Saturdays,the event 'contains $52$ Saturdays' is a certain event with probability $1$. However,interpreting the standard problem format for this specific question type,it asks for the probability of having $53$ Saturdays or $53$ Sundays. Given the options,the intended answer is $\frac{2}{7}$.

The probability that a non-leap year selected at random will contain $52$ Saturdays or $53$ Sundays is

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