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

• 5

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.

• 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

hope it helps..
• -6

• 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
• -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

• -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
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
• 52
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
• -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

• 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

• 4
May it helps youu.....

• 7

• 1

• 0

• 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

• 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
• 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
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