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(C) leap year consists of $366$ days,which is equivalent to $52$ weeks and $2$ extra days.The $7$ possible pairs for these $2$ extra days are:$(i)$ (S

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$\frac{3}{7}$

Vedclass · Answer

(C) leap year consists of $366$ days,which is equivalent to $52$ weeks and $2$ extra days.The $7$ possible pairs for these $2$ extra days are:$(i)$ (Sunday,Monday),$(ii)$ (Monday,Tuesday),$(iii)$ (Tuesday,Wednesday),$(iv)$ (Wednesday,Thursday),$(v)$ (Thursday,Friday),$(vi)$ (Friday,Saturday),$(vii)$ (Saturday,Sunday).Let $A$ be the event that the leap year contains $53$ Sundays,and $B$ be the event that it contains $53$ Mondays.From the list above,the pairs containing Sunday are $(i)$ and $(vii)$,so $P(A) = \frac{2}{7}$.The pairs containing Monday are $(i)$ and $(ii)$,so $P(B) = \frac{2}{7}$.The pair containing both Sunday and Monday is $(i)$,so $P(A \cap B) = \frac{1}{7}$.Using the addition rule for probability,the probability of $A$ or $B$ is:$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{2}{7} + \frac{2}{7} - \frac{1}{7} = \frac{3}{7}$.

The probability that a leap year selected at random contains either $53$ Sundays or $53$ Mondays is

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