What is the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays in a non–leap year ?

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A non-leap year consists of 365 days. Therefore in a non-leap year there are 52 complete weeks and 1 day over which can be one of the seven days of the week.

Possible outcomes n(S) = 7 = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}.

∴ Number of possible outcomes n(S) = 7

(As there are seven days in a week)

Let A : Getting the extra day as Sunday or Tuesday or Thursday

⇒ A = {Sunday, Tuesday, Thursday}

⇒ n(A) = 3

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{3}{7}.\)