If a leap year is selected at random,what is | vedclass

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(B) leap year consists of $366$ days,which is equal to $52$ weeks and $2$ extra days.In $52$ weeks,there are exactly $52$ Tuesdays.To have $53$ Tuesda

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$\frac{2}{7}$

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(B) leap year consists of $366$ days,which is equal to $52$ weeks and $2$ extra days.In $52$ weeks,there are exactly $52$ Tuesdays.To have $53$ Tuesdays,one of the remaining $2$ days must be a Tuesday.The possible pairs for the $2$ extra days are:(Monday,Tuesday),(Tuesday,Wednesday),(Wednesday,Thursday),(Thursday,Friday),(Friday,Saturday),(Saturday,Sunday),(Sunday,Monday).Total number of possible outcomes $= 7$.The favorable outcomes where at least one day is a Tuesday are (Monday,Tuesday) and (Tuesday,Wednesday).Number of favorable outcomes $= 2$.Therefore,the probability $= \frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{2}{7}$.

If a leap year is selected at random,what is the chance that it will contain $53$ Tuesdays?

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