What is the probability of getting 52 Sundays in a :-
1) Leap Year
2) Non Leap Year
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Answers
In a non-leap year there will be 52 Sundays and 1day will be left.
This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday,friday,Saturday, Sunday.
Of these total 7 outcomes, the favourable outcomes are 1.
Hence,
the probability of getting 53 sundays = 1 / 7.
and,
From the above,
A leap year has 366 days or 52 weeks and 2 odd days.
The two odd days can be {Sunday,Monday},{Monday,Tuesday},{Tuesday,Wednesday},{Wednesday,Thursday},{Thursday,Friday},{Friday,Saturday},{Saturday,Sunday}.
So there are 7 possibilities out of which 2 have a Sunday.
So the probability of 53 Sundays is 2/7.
hope it helps.
Hello!!
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⭕Solution:
1. Leap Year.
➡️ Earlier I also thought , it should be 1 but more understanding and practising the questions on probability , my concept became clearer.
➡️ A leap year has 366 days = 52 × 7 + 2 .
➡️ So, there are 52 weeks i.e. 52 sundays but the two extra days changes the game.
➡️ Now , we have to focus on these 2 days .
➡️ The extra 2 days is the main concept to understand = (sunday, monday) , (monday-tuesday) , (tuesday-wednesday) ,(wednesday-thursday) ,(thursday-friday) , (firday-saturday), (saturday-sunday)
➡️ If any of the two days is SUNDAY , then total number of favorable outcomes when SUNDAYS are not exactly 52 becomes 2.
➡️ Probability of not exactly 52 sundays = 2 / 7
➡️ Therefore, Probability of exactly 52
sundays = 1 - 2 /7 = 5/7
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2. Non Leap Year.
➡️ A non leap year has 365 days, so, we can have 52 weeks (52 x 7=364 days) and we're left with one day extra (365–364=1).
➡️ This day can be any one of the seven different days and since we already had 52 Sundays because we already completed 52 weeks, we don't want this extra day to be a Sunday.
➡️ So, the no. of favorable days are 6 and total possibilities are 7.
➡️ This the answer is 6/7.
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Thanks!!