What is the probability of getting 53 Fridays in a leap year?
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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 clear with it OK bye
Answered by
168
A leap year consists of 366 days.
52 weeks + 2 days.
These 2 days can be: (mon,tue),(tue,wed),(wed,thu),(thu,fri),(fri,sat)(sat,sun)(sun,mon)
Thus, the total number of cases = 7.
The number of cases in which we get Friday = 2(Thu, Fri)(Fri, Sat).
Therefore the required probability = 2/7.
Hope this helps!
52 weeks + 2 days.
These 2 days can be: (mon,tue),(tue,wed),(wed,thu),(thu,fri),(fri,sat)(sat,sun)(sun,mon)
Thus, the total number of cases = 7.
The number of cases in which we get Friday = 2(Thu, Fri)(Fri, Sat).
Therefore the required probability = 2/7.
Hope this helps!
siddhartharao77:
Thanks for the brainliest Rahi
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