Find the probability that a non leap year has 53 sundays..

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G Prasad

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As there are 52 weeks in a year. 52 weeks correspond to 7*52=364 days. So, the last day is the one which should be a Sunday for you to get 53 Sundays. Sample space has seven days as options. So probability of getting 53 Sundays in a non-leap year is 1/7
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SME in Mathematics & Statistics

A non leap year has 365 days. There are 52 weeks each having 7 days. This amounts to 52 x 7 = 364 days. The remaining 1 day can be any day among Monday, Tuesday,..., Sunday. Hence the required probability is 1/7.
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Pursuing my IAS goal since 2010-11

Divide 365 by 7, we get 1 remainder. 365 days has 52 weeks and a extra day which can be any day among mon,tue,wed....sun. So 53rd sunday probability=1/7.
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Probability of getting 53 Sundays in a non leap year = 1/7
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Spoken English, BTech Tuition, C Language, C++ Language, CAD etc., with 4 years of experience

52 weeks*7 days per week = 364 days The remaining day in case of a non leap year is 1 (365-364) The possibilities of this day are Sun, Mon, Tue, Wed, Thu, Fri, Sat = 7 The favorable outcome = Sun = 1 Therefore the probability of getting 53 sundays in a non leap year = 1/7
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1/7
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Headstart to assured 95+ score in math

1/7 i think.
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1/7
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1/7
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Online Tutor-Mathematics, Science, Competitive Exam Preparation

1/7 For non-leap year. 1st of Jan must be Sunday. 2/7 for a leap year. If 1st Jan is either Saturday or Sunday.
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