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What is the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays in a non–leap year ? |
Answer» 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) = \(\frac{n(A)}{n(S)}\) = \(\frac{3}{7}.\) |
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