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(C) leap year consists of $366$ days,which is $52$ weeks and $2$ extra days. The $7$ possible outcomes for these $2$ extra days are:$(i)$ (Sunday,Mond

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(C) leap year consists of $366$ days,which is $52$ weeks and $2$ extra days. The $7$ possible outcomes 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 has $53$ Sundays,and $B$ be the event that it has $53$ Mondays.Then $P(A) = 2/7$,$P(B) = 2/7$,and $P(A \cap B) = 1/7$ (since the only case with both is Sunday and Monday).The required probability is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$P(A \cup B) = 2/7 + 2/7 - 1/7 = 3/7$.

What is the probability that a randomly selected leap year has $53$ Sundays or $53$ Mondays (in $/7$)?

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