The probability that a leap year will have 53 | vedclass

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(B) leap year has $366$ days,which consists of $52$ weeks and $2$ extra days.The possible combinations for these $2$ days are: (Sunday,Monday),(Monday

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$\frac{3}{7}$

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(B) leap year has $366$ days,which consists of $52$ weeks and $2$ extra days.The possible combinations for these $2$ days are: (Sunday,Monday),(Monday,Tuesday),(Tuesday,Wednesday),(Wednesday,Thursday),(Thursday,Friday),(Friday,Saturday),and (Saturday,Sunday).There are $7$ total possible outcomes.Let $A$ be the event of having $53$ Fridays and $B$ be the event of having $53$ Saturdays.$P(A) = \frac{2}{7}$ (since Friday occurs in (Thursday,Friday) and (Friday,Saturday)).$P(B) = \frac{2}{7}$ (since Saturday occurs in (Friday,Saturday) and (Saturday,Sunday)).$P(A \cap B) = \frac{1}{7}$ (since (Friday,Saturday) is the only common outcome).Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$P(A \cup B) = \frac{2}{7} + \frac{2}{7} - \frac{1}{7} = \frac{3}{7}$.

The probability that a leap year will have $53$ Fridays or $53$ Saturdays is

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