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Find the probability of having 53 Fridays in a -year which is not a leap year. |
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Answer» A year which is not a leap year is Normal year having 365 days. So, the sample space for 1 extra day is expressed as follows: U = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday} ∴ Total primary outcomes n = 7 A = Event that having 53 Fridays in a year which is not a leap year.= {Friday} ∴ Favourable outcome for the event A is m = 1 Hence, P(A) = \(\frac{m}{n} = \frac{1}{7}\) |
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