what is the probability that a leap year has 53 sundays and 53 mondays?

 53 366  and 53 366

  • -9

 53/ 366 and 53/ 366

  • -11

it will be  2/7 and 2/7

  • -1
  • leap year has 366 days which is 52 weeks + 2 days there can be 53 sundays only when the last 2 days are
  • (saturday , sunday or sunday monday) so P(53 sundays)= 2/7 as total there can be 7 cases similarly P(53 mondays) = 2/7 P(53 sundays or 53 mondays)= P(53 sundays union 53 mondays)
  • = P(53 sundays)+ P(53 mondays) - P(53 sundays and 53 mondays) which is seen in only 1 case so P(53 sundays or 53 mondays)
  • = 2/7+2/7-1/7 = 3/7
  • 10

2/7

  • 2

366 days = 52 weeks + 2 days

2 days can be MT, TW, WT, TF, SS , SM =7

P=2/7

HOPE U UNDERSTAND .................

  • 1
for a NON leap year: No of days = 365 
No of week = 52Week +1 day 
so first 52weeks will surely have a monday. 
Event that a Monday will come on the remaining 1 day: n(E) = 1 [E= {Monday}] 
Sample Space (no of Possibilities) n(S) = 7 [ S= {M,T,W,T,F,S,S} ] 
Therefore P= n(E)/n(S) = 1/7 


for a Non leap year: No of days = 366 
No of week = 52Week + 2 days 
so first 52weeks will surely have a monday. 
Event that a Monday will come on remaining 2 days: n(E) = 2 [E = { (Sun,Mon) / (Mon,Tue) } ] 
Sample Space (no of Possibilities) n(S) = 7 
[S= { (Sun,Mon) (Mon,Tue) (Tue,Wed) (Wed,Thu) (Thu,Fri) (Fri,Sat) (Sat,Sun) } 
Therefore P= n(E)/n(S) = 2/7..  
  • 37
2/7 is the only answer
  • -3
A leap year has 366 days.
when 366 is divided by 7,you get 52 as quotient and 2 as remainder.
Those 2 days can be MT,TW,WTh,ThF,FS,SSu,orSuM
​No of favourable outcomes=2
​So,P(E)=2/7
 
  • 19
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
  • 5
366 days ➗ 7days = 52 weeks and 2 days : These two days can be Sunday monday Monday tuesday Tuesday wednesday Wednesday Thursday Thursday Friday Friday Saturday Saturday Sunday And therefore probability is 2/7
  • 0
In a leap year ,total number of days =366 In 366 days there are 52 weeks and 2 days Now two days may be Sun and Mon Mon andTue Tue and Wed Wed andThur Thur and Fri Fri and Sat Sat and Sun Now, in total 7 probabilities , Sun and Mon come together in one time So, and=1/7
  • -2
Hope you understand it,

  • 7
2/7
  • 0
2/ 7
  • 0
A leap year has 366 days
So no: of weeks = 52
These 52 weeks surely have 52 sundays and 52 mondays

No: of remaining days = 2

These two days could be :
{(Sun,Mon),  (Mon,Tue), (Tue, Wed), (Wed,Thur), (Thur,Fri), (Fri,Sat), (Sat,Sun)}

No: of favourable outcomes (i.e., the remaining 2 days are Sunday and Monday to make it 53 sundays and 53 mondays) = 1

Total possible ways = 7

Therefore, P(E) = 1/7
 
  • -1
1/7
1/7
  • 1
ONLY AND ONLY THE THE CORRECT ANSWER IS 2/7
  • 0
Please see answer below
  • 0
2/7 probablity
  • 0
Probability of 53 Sundays are 2/7 and 53 Mondays are 2/7
  • 0
Probability is 2/7
  • 0
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  • 0
In a leap year the no of days is 366, right. And we also know that in a year(365) there are 52weeks.
But there is one day extra in a leap year.
So that day can be any day of the week( sunday, monday, tuesday, wednesday, thursday, friday or saturday)
In a normal year we have 52mondays right.
The probability of having 53mondays in a leap year is 1/7
Because that one extra day may be any day of the week
  • 0
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  • 0
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