Probability that a leap year will have 53 sunday?explain, Mathematics

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A leap year has 366 days, therefore 52 weeks i.e. 52 Sunday and 2 days. The remaining 2 days may be any of the following : (i) Sunday and Monday (ii) Monday and Tuesday (iii) Tuesday and Wednesday (iv) Wednesday and Thursday (v) Thursday and Friday (vi) Friday and Saturday (vii) Saturday and Sunday For having 53 Sundays in a year, one of the remaining 2 days must be a Sunday. n(S) = 7 n(E) = 2 P(E) = n(E) / n(S) = 2 / 7

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3/7/2013 5:13:25 AM Sol. A non - leap year contains 365 days 52 weeks and 1 day more. i) We get 53 Sundays when the remaining day is Sunday. Number of days in week = 7 ∴ n(S) = 7 Number of ways getting 53 Sundays. n(E) = 1 n E 1 n S 7 = ∴ Probability of getting 53 Sundays =1/7
3/7/2013 5:13:41 AM Sol. A non - leap year contains 365 days 52 weeks and 1 day more. i) We get 53 Sundays when the remaining day is Sunday. Number of days in week = 7 ∴ n(S) = 7 Number of ways getting 53 Sundays. n(E) = 1 n E 1 n S 7 = ∴ Probability of getting 53 Sundays =1/7

3/7/2013 5:13:25 AM

Sol. A non - leap year contains 365 days 52 weeks and 1 day more.
i) We get 53 Sundays when the remaining day is Sunday.
Number of days in week = 7
∴ n(S) = 7
Number of ways getting 53 Sundays.
n(E) = 1
n E 1
n S 7
=
∴ Probability of getting 53 Sundays =1/7

3/7/2013 5:13:41 AM

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