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what is the probability that a non leap has 53 sundays?

Your answer for both non leap and leap year.

  • 5
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 Normal year has 52 weeks (365/7) + 1 extra day ie. 52 Sundays + 1 day which can be any of the seven days So probability of 53 Sundays === Probability of that day being a sunday = 1/7
OR
The year MUST start on a Sunday, so you have only one favorable chance out of a total of 7 possibilities - a probability of 1/7.

  • 133

1:7 is the probability of getting 53 sundays in a non leap year..Like 2011 started frm saturday, that means 2011 is having 53 saturdays.If this year started frm sunday then this year is consisting 53 sundays......

HOPE THIS HELPS YOU

  • -7

 Its 1/7

  • -13

Normal year has 52 weeks (365/7) + 1 extra day ie. 52 Sundays + 1 day which can be any of the seven days So probability of 53 Sundays === Probability of that day being a sunday = 1/7
OR
The year MUST start on a Sunday, so you have only one favorable chance out of a total of 7 possibilities - a probability of 1/7.

SO PLEASE CLICK THUMBS UP AT RIGHT BOTTOM OF THIS PAGE NAA...
 


 

  • 30

thank you everyone! :)

  • -7

1/7

  • -9

Normal year has 52 weeks (365/7) + 1 extra day ie. 52 Sundays + 1 day which can be any of the seven days So probability of 53 Sundays === Probability of that day being a sunday = 1/7

  • 4

Total number of days in a leep year = 366

Total number of complete week in leap year = 52

Remaining days = 2

They would be ,

Mon, Tue

Tue, Wed

Wed, Thu

Thu, Fri

Fri, Sat

Sat, Sun

Sun, Mon

Total number of outcome = 7

Total number required outcome =2

Therefore, Required Probablity = 2/7.....!!!!.....

  • 13

thanx everyone! this really helped me!

  • -4
Its 1/7
  • -9
its 2/7
  • -6
1/7
  • -7
1/7 or 2/7
  • -8
p(of getting a sunday in a leap year) = 1 / 7

hope it helps..
  • -7
the answer is 2/ 7 

hope it helps..
  • -6
Please find this answer

  • 35
Its 1/7
  • -5
hey gooooood  question  ans.1/7
​

 
  • -6
1/7
 
  • -3
The leap year has 366 no. of days. the sunday comes 52 times and 2 days are left. those can be   
    mon,tue
tue,wed
wed,thur
thur,fri
fri,sat
sat,sun,
sun,mon

so there are 7 no. of outcomes so probablity of getting sunday is 2/7
  • -8
2/7

 
  • -6
Its 1/7
  • -8
1/7
 
  • -7
p[53 sunday=1/7
  • -5
1/7
  • -7
Answer is 53/365
  • -8
NO OF DAYS IN A WEEK=7
SUNDAY  IS NO OF THE WEEKDAYS= 1
P(THE LEAP DAY IS SUNDAY)=1/7
  • 1
1/7
  • -6
Prime 7/17 Divisibleby 2 and 3 both 2/17
  • 3
This answer in photo

  • -5
A non-leap year is one which has 365 days(365.25 to be exact but for calculations we will take it to be 365)
So, let us calculate the no. of weeks in an non leap year(it is the same for a leap year, but we actually want the remainder as we will see later)
7)365(52
  35
(-)_____
 15
 14
(-)__
   1

this means that there are 52 weeks(and hence 52 Sundays) in a non leap year and one extra day. Out of seven days, the probability that that extra day is Sunday=(no.of Sundays in a week)/Total no of days in the week
=1/7
Hence, the probability that a non leap  year has 53 Sundays is 1/7
  • -5
1/7 
  • -3
The probability is 1/7 since there are 365 days( 364+1) days and the 1extra day could be any day from the seven days in a week . Hence this 1 day could be a sunday as well.
  • 7
1/7
  • -5
1/7
  • -6
in normal year there were 365 days in which 52 weeks and 1 odd day .the probability of getting 1 sunday equal to 1/7 because total no of outcomes equal to no of days in a week.
  • -1
1/7
 
  • 3
Here is the answer to your query
      number of days in a non-leap year=365 days = 52 weeks and 1 day
       Each week there is 1 Sunday , but here there is 1 day extra

     That 1 day can be either{Sunday, monday, Tuesday, Wednesday, Thursday, Friday, Saturday} 
        Therefore,Probability of getting that day a Sunday=1/7

Thus,Probability of getting 52 Sundays= 1,whereas getting 53 Sundays =1/7

HOPE IT HELPED YOU
THUMBS UP PLEASE...
  • 51
A fair dice is role . Probability of getting a no. x such that 1
  • 4
IN AN NON-LEAP YEAR
NO. OF DAYS ARE = 366 DAYS 
NO. OF WEEKS ARE = 52 WEEKS + 1 DAY (ANY DAY OF A WEEK)
THEREFORE, P (Q GETTING 53 SUNDAYS IN AN NON-LEAP YEAR) = 1/7
  • 7
Wrong Anwer
  • -6
1/7 because normal year have 52 weeks + one day than 52+1 sunday so.
  • -6
hope it will helpfull
  • -3
1/7
 
  • 2
1/7
  • -4
pobabilty of this is 1/7 thanks for all the reader
  • 1
1/7
  • -3
1/7
  • -4
Leap year has 366 days
i.e 52weeks and 2days
Therefore, Sunday comes 52 times and 2 days is chance of probability 
These two days can be either the following (sample space):
      Sunday, Monday
      Monday, Tuesday 
      Tuesday, Wednesday
      Wednesday, Thursday
      Thursday, Friday
      Friday, Saturday
      Saturday, Sunday 
n(s):- 7
We saw that "Sunday" comes twice. So,
P(getting Sunday):- 2/7

GIVE IT A THUMBS UP GUYS!!!

 
  • -1
1/7
  • 6
1/7
  • 6
2/7
  • 1
Year must end on a Sunday is it not? Since it is 52*7+1 . The one extra day is Sunday..
  • 2
Hope it help you.

  • 8
There are 365 days in a year, i.e., 52 weeks and 1 day extra......(365/7) Each week has a Sunday.... Now that one day can be any of the seven days......i.e., Sunday to Saturday...... Thus, the probability of getting that day Sunday will be 1/7
  • 5
its 2/7
 
  • 2
1/7
  • 4
1/7
 
  • 2
1/7
  • 4
A non leap year has 365 days.
i.e., 52 weeks and 1 day which has possibility of being any of the 7 days.
Therefore, probability that a non leap year having 53 Sundays is 1/7.
  • 5
53/365
  • 2
1/7 is the answer
 
  • 4
May it helps youu.....

  • 7
Please find this answer

  • 1
This is your answer to the question. He this helps....

  • 0
This is your answer to the question.

  • 0
1/7 is the probability
  • 4
1/7                                        
  • 2
1/7 as 52 week are proper means there is 52 Sundays 52mondays etc but 1 day is extra of a week that can be Sunday . Monday or etc so probability is what 1/7
  • 0
translate
  • 2
1/7 is theprobability
  • 2

It’s 1/7.

In any year,there’re atleast 52 weeks which accounts for 52*7=364 days.Now a non-leap year has 365 days i.e. 1 day extra.For this day possibilities will be {mon ,tue, wed, thu, fri, sat, sun}.

Thus total no. Of possibilities are 7,out of which there’s one possibility of getting 53 Sundays i.e. when the extra day is Sunday.

Now,

probability =total no. Of favourable outcomes/total possible outcomes

So,probability=1/7.

  • 0
Please find this answer

  • 0
its 0,as none of the year has 53 weeks i guess
 
  • 2
1/7
 
  • 0
No of weeks=52 
it is 52 weeks in 364 days and one more day is left.
P(Sunday)=1/7
Therefore P(53 Sundays in a year)=1/7
  • 2
53 /365
  • 2
answer is 2/7
  • 0
Total no. days = 7
Since 7?52=364
Therefore ,
Favourable condition =1
Therefore
P(53 sundays ) = 1/7
  • 2
No of days in leap year=366
366 days = 52 weeks alnd 2 days
So there are 53 sundays in leap year
Remaining 2 days can be-
1)sunday and monday
2)monday and tuesday
3)tuesday and wednesday
4)wednesday and thursday
5)thursday and friday
6)friday and saturday
7)saturday and sunday

So;no. Of events= 7
No.of favourable events=2
P(E)= 2/7
  • 0
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.
  • -1
In a leap p(53sundays)=2/7 so
In non leap year p(53sundays)=2/6
  • 4
2/7 is the answer...
  • 2
The probability is 1/7
  • 0
?Normal year has 52 weeks (365/7) + 1 extra day ie. 52 Sundays + 1 day which can be any of the seven days So probability of 53 Sundays === Probability of that day being a sunday = 1/7
OR
The year MUST start on a Sunday, so you have only one favorable chance out of a total of 7 possibilities - a probability of 1/7.
  • 2
D.r apj abdul kalam
  • -1
Non leap year contain = 365 days. Total Sunday = 53. Probability Sunday = 53/365.
  • 0
In a non leap year there will be 52 sundays and 1 day will be left. This 1 day can be Sunday, Monday, Tuesday,Wednesday, Thursday, Friday, Saturday. Total outcomes 7 and favourable numbers are 1. Hence the probability of getting 53 sundays = 1/7
  • 0
Check it out !!

  • 0
non leap year = 365 days
Weeks = 52 weeks
1 day is extra
That one day can be ,
A = {Mon , Tue , wed , Thur , Fri , sat , sun}
N(s) = 7
If the day should be Sunday then
N(a) = 1
P(e) = N(a)/N(s)
P(e) = 1/7
  • 0
Please find this answer

  • 0
As there are 365 days in a leap year. No.of weeks=52 weeks.
No. Of days left is 1.
1 can be mon, tue, wed, thurs, fri, sat, sun
Total no.of fays=7
P(53 sundays) is 1/7.
  • 0
P( an ordinary year of 53 Sundays) =1/7
  • 0
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