what is da probability that a leap year have 53 tuesdays?

want explanation ?
 

thnx..... bt i didnt get dat..... :|

See yaar 

There are 366 days in a leap year !

1 tuesday in 1 week 

Number weeks = 52 weeks ...
 
but 2 days left 


52 weeks so 52 tuesdays

Remaining 2 days ..derz a possibility of getting tuesday on the 365th day or 366th day 

But there are 7 days  ( sun mon tues wed thu fri sat ) in a week 

The two days can be in any combination as 

sun , mon

mon , tues

tues wed 
 etc etc 


The last 2 days can be any consequtive days


we want tuesday  The possible outcomes are either the 1st day as tuesday or next day 

The two combinations

Mon tue   nd Tue wed

contains tue ..

So 2 favourable outcomes  der are  totally 7 outcomes 

so The probablity = 2 / 7

Any doubtzzz

thnk u......:)...

 :)) Keep Smiling !

:)...

There are 52 weeks and 2 days in a leap year

Therefore,

52 * 7 = 364

Remainder = 2

Therefore the probability of getting a Tuesday in those days would be 2/7

Since there are 366 days in a leap year .
Number of tuesdays = 52 ( As 52 x 7 = 364 )
So , for the leap year we get two more days .
Now , we can get 53 tuesdays . if the two days are either monday nd tuesday or tuesday nd wednesday

Therefore required Probability of 53 tuesdays = 2/7

There are 53 Tuesdays in a leap year if it starts on a Monday or Tuesday.

This can't happen in 2/7 of all leap years, because the distribution of leap years and first-days-of-the year repeats every 400 years. There are 97 leap years in each such period, which is not a multiple of 7.

More precisely, in every 400-year period,

• 13 leap years start on a Monday
• 14 leap years start on a Tuesday
• 14 leap years start on a Wednesday
• 13 leap years start on a Thursday
• 15 leap years start on a Friday
• 13 leap years start on a Saturday
• 15 leap years start on a Sunday

So the probablility that a leap year chosen uniformly among the leap years in a cycle starts on a Monday or Tuesday (and so contains 53 Tuesdays) is 
 
    (13+14)/97= 27/97