1.

Find the probability of having 53 Fridays in a -year which is not a leap year.

Answer»

A year which is not a leap year is Normal year having 365 days.
There are 52 weeks, i.e., 52 × 7 = 364 days and 1 day extra in a normal year.

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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