IIT JEE , AIEEE Forum - Mathematics , Algebra

shine

a question of prabability

ONE DATE IS SELECTED AT RANDOM FROM ALL THE DATES OF 21st CENTURY i.e. from 01.01.2001 to 31.12.2100 WHICH ARE WRITTEN IN THE FORM OF 'DDMMYY'.
what is the probability that it is a PELINDROME date?

[a PELINDROME DATE appears similar in both the ways 'DDMMYY' & 'YYMMDD' e.g. 10.11.01]

shine

come on !
someone reply plzzzzzzzzzz!!!!!!!!!!!!

KAB

Is the ans 30/36520?????????

KAB

Total no of outcomes(dates) =365*75+366*25=36525 [Since there are 25 leap years.]

To calculate required outcomes:

The month 11 is the only possible month.The no. of days are 30.

When day is 01 the year should be 10.

When day is 02 the year should be 20.

Similarly when the day is 30 the year should be 03.

So the required probability is 30/36525.

PS:I'm not sure that this is correct.....

shine

hay kab
actually i dont have the answer
but i m also getting the same favourable days
however dont u think tgere should b 25 leapyears in 100 years
just think!!!!!

KAB

Sorry!!!!Just a silly mistake.......

magiclko

hey adarsh m afraid u r wrong.... actually the question seems more of PC and less of probability....newayz here's the solution

For the date to be palindrome and belonging to year s [2001, 2100]...only possible thing is , that the month has to be feburary.... i.e. in the following format
_ _ 0 2 | 2 0 _ _
a b c d

now a can be either 0 or 1 or 2 and b can be frm [1,8] for a non-leap year...and [1,9] for a leap year...

as the date has to be palindrome d aslo has to be in either 0, 1 or 2
case 1: when year is a leap year...
then d can be either 0 or 2....
if d=2,
2 _ 0 2 | 2 0 _ 2
b c

then for year to be leap c (and also b) belongs {1,3,5,7,9} that is 5 cases

if d=0
0 _ 0 2 | 2 0 _ 0
b c
then for year to be leap c (and also b) belongs {2,4,6,8} that is 4 cases

case 2: when year is not a leap year....
if d=2,
2 _ 0 2 | 2 0 _ 2
b c
now b (and c) can be any one of {0,2,4,6,8}...i.e. 5 cases

if d=1,
1 _ 0 2 | 2 0 _ 1
b c
then b (and c) can be any of the [0,8], i.e. 9 cases

if d=0,
0 _ 0 2 | 2 0 _ 0
b c
then b (and c) can be any of the [1,8], i.e. 8 cases

So, the total no of favourable cases = sum of all this = 31

and total no of days frm 1st jan 2001 to 31st december 2100 = x (calculate it plsss

)

so the probabilty = 31/x

phew the question seems bit lengthy

... but i cant think of anything else... any short approach is most welcomed!!!

magiclko

ooooooooooops i did a big mistake.... i took the date in DDMMYYYY format .... the whole question went wrong....

newayz the approach will be the same and i think KAB is right, except his silly mistake!!!!

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