what is the probability that a non leap year may cointain 53 Sunday
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the probability that a non leap year may contain 53 Sundays is 1/7
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Heya,
Non-Leap year = 365 days
1 week = 7 days
365/7
=> 52 (Quotient) and 1 (Remainder)
Therefore,
1 day maybe:-
(Monday) , (Tuesday) , (Wednesday) , (Thursday) , (Friday) , (Saturday) , (Sunday) = 1/7
Hence,
P (53 Sundays) = 1/7
Hope my answer helps you :)
Regards,
Shobana
Non-Leap year = 365 days
1 week = 7 days
365/7
=> 52 (Quotient) and 1 (Remainder)
Therefore,
1 day maybe:-
(Monday) , (Tuesday) , (Wednesday) , (Thursday) , (Friday) , (Saturday) , (Sunday) = 1/7
Hence,
P (53 Sundays) = 1/7
Hope my answer helps you :)
Regards,
Shobana
paviter1:
thanks dear
My pleasure :)
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