What is the probability that a non-leap year | vedclass

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(C) non-leap year consists of $365$ days.Dividing $365$ by $7$,we get $52$ weeks and $1$ extra day.This extra day can be any one of the following: {Su

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$\frac{1}{7}$

Vedclass · Answer

(C) non-leap year consists of $365$ days.Dividing $365$ by $7$,we get $52$ weeks and $1$ extra day.This extra day can be any one of the following: {Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday}.Thus,the total number of outcomes is $n(S) = 7$.For the year to have $53$ Wednesdays,the extra day must be a Wednesday.There is only $1$ favorable outcome,which is {Wednesday}.Thus,$n(E) = 1$.The probability $P(E) = \frac{n(E)}{n(S)} = \frac{1}{7}$.

What is the probability that a non-leap year has $53$ Wednesdays?

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