Math Is Fun Forum

soha

Real Member

Registered: 2006-07-07

Posts: 2,530

........help needed urgently

i have wasted 15 mins for this prob, now its ur chance .what is the probability of getting 53 sundays in ayear.pls try & help......................

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LQ

Real Member

Registered: 2006-12-04

Posts: 1,285

Re: ........help needed urgently

You want number of sundays as a function of days?
sundayless days in first seven days = q (q(max) = 6)
days in a year = days
1+ (days-(q+1))/7

This is a matrix (a mathematical matrix, that is). If (days-(q+1))/7 ≥ 52, then you have 53 sundays

for days = 365, the chance is 1/7*100%
for days = 366, the chance is 2/7*100%
the chance for days to be 366 is 1/4
the chance for days to be 365 is 3/4
(1/4*2/7 + 3/4*1/7)*100 = 17,857% (=1/5.6 or 5/28)

Last edited by LQ (2006-12-10 05:17:10)

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I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...

Toast

Real Member

Registered: 2006-10-08

Posts: 1,321

Re: ........help needed urgently

How do you get from your matrix to the probabilities of 1/7 and 2/7?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: ........help needed urgently

I deleted your previous post for you, Toast. I'm not sure what the problem was, but it's solved now, anyway.

A possibly easier way of finding the number of Sundays in a year would be to consider a year with 364 days. 364/7 = 52, so that year would always have exactly 52 sundays in it, no matter what day it started on.

Therefore, a year with 365 days will have 52 complete weeks and one extra day, so it will have 53 sundays if the first day was a sunday, or 52 otherwise. So, a 1/7 chance of having 53 sundays.

Similarly, a year with 366 days in it will have a 2/7 chance.

And the rest of the calculation is the same.

Of course, if you wanted to be really specific, you could note that leap years are "unincluded" if they are multiples of 100, but then "reincluded" if they are multiples of 400.

So the chance of any year being leapy is actually not 1/4, but 97/400.

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LQ

Real Member

Registered: 2006-12-04

Posts: 1,285

Re: ........help needed urgently

I 2 used the simple trick that (days-(q+1))/7 can be ≥ 52 for 2 values of  q, 0 and 1, and q can only have 7 values, 0,1,2,3,4,5,6. So in 2 of seven cases every fourth year the equation yeal ≥ 52, and the other 3 years, only one value of q (0) makes the equation yeal ≥ 52.

And I figured that someone would speak about leapy being 97/400 chance or so. Normaly, such things are not taken into consideration in 7:nth grade though. Yah know.

And i think there is no conspiracy about it all really. This we wrote should work.

Last edited by LQ (2006-12-10 06:33:45)

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I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...