Find the probability that a leap year contains 53 fridays
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Answered by
15
A leap year consists of 366 days, i.e., 52 complete weeks and 2 days remain. These 2 days can be any of the following:
(Mon, Tues), (Tues, Wed), (Wed, Thur), (Thur, Fri), (Fri, Sat), (Sat, Sun)
Thus, total number of cases = 7
Number of cases in which we get a Friday = 2
∴Probability of getting 53 Fridays in a leap year = 2/7
Answered by
3
There are 52 weeks and 2 days in a leap year so there are 52 Fridays in a leap year and last 2 days can be any 2 fays(mon and Tues, Tues and wed........ ) as 2 events out of 7 have 'Friday' so probability is 2/7
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