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