The probability that a leap year will have 53 sunday is ? and how please explain it ?(a)1/7 (b) 2/7 (c) 5/7 (d)6/7

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

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

Approved

a leap year consits  of 366 days in those 364 days are completely 52 weeks so it contains 52 sundays remaining 2 days there are 2/7 probabilities are there.

ans is b.

in leap yr der r 2 odd daysso probability for sunday to be one of them is 2 out of 7 as it may be Sunday,Monday or Saturday,Sunday

A leape year consists of 366 days Normal year consists of 365 days One year consists of 52 weeks One week consists of 7 days52×7= 36452 Sundays per year Extra 2 days So probability =2/7

(monday,sunday),(Monday,tuesday),(Tuesday,wednesday)(Wednesday,thursday),(Friday,saturday)(Saturday,sunday). Possible outcome=2Total number of outcomes=7P(E)=2/7

In a Leap Year, there are 366 Days.366 Days = 52 Weeks and 2 DaysThus, a Leap Year has always 52 Sundays.The Remaining 2 Days can be :(i) Sunday & Monday ; (ii) Mon & Tue ; (iii) Tue & Wed ; (iv) Wed & Thu ; (v) Thu & Fri ; (vi) Fri & Sat ; (vii) Sat & Sun. Clearly there are 7 Elementary events associated. Let A be the event that a Leap Year has 53 Sundays.Clearly, the event A will happen if the last two days of the leap year are either Sun & Mon or Sat & Sun. Favourable Outcomes = 2Required Probability = 2/7.