The probability that a non-leap year selected | vedclass

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Question

(B) non-leap year has $365$ days, which is equal to $52$ weeks and $1$ extra day. The sample space for this extra day is $S = \{ 	ext{Sunday, Monday,

Vedclass · Accepted Answer

$\frac{2}{7}$

Vedclass · Answer

(B) non-leap year has $365$ days, which is equal to $52$ weeks and $1$ extra day. The sample space for this extra day is $S = \{ 	ext{Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday} \}$. There are $7$ possible outcomes for this extra day. Let $A$ be the event of having $53$ Sundays and $B$ be the event of having $53$ Saturdays. The year will have $53$ Sundays if the extra day is a Sunday, so $P(A) = \frac{1}{7}$. The year will have $53$ Saturdays if the extra day is a Saturday, so $P(B) = \frac{1}{7}$. Since these events are mutually exclusive, the probability of having $53$ Sundays or $53$ Saturdays is $P(A \cup B) = P(A) + P(B) = \frac{1}{7} + \frac{1}{7} = \frac{2}{7}$.

The probability that a non-leap year selected at random will have $53$ Sundays or $53$ Saturdays is

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