The probability of a non-leap year having 53 | vedclass

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(D) non-leap year has $365$ days.$365$ days $= 52$ weeks and $1$ extra day.$52$ weeks contain $52$ Fridays.For the year to have $53$ Fridays,the remai

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$\frac{1}{7}$

Vedclass · Answer

(D) non-leap year has $365$ days.$365$ days $= 52$ weeks and $1$ extra day.$52$ weeks contain $52$ Fridays.For the year to have $53$ Fridays,the remaining $1$ extra day must be a Friday.The possible outcomes for the extra day are: {Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday}.Total number of outcomes $= 7$.Number of favorable outcomes (Friday) $= 1$.Therefore,the probability $= \frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{1}{7}$.

The probability of a non-leap year having $53$ Fridays is $\ldots \ldots \ldots$

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