Question 23954: ADDED ADDITIONAL EXPLANATION AND WORKING FOR THE SECOND PROBLEM..ONLY YOU NEED TO DO THE ARITHMATIC
VENUGOPAL
OH!THIS IS A RESEARCH PROJECT FOR YOU ..THEN IT IS OK ..
THEN LET US ANALYSE A BIT..THE FIRST PART IS EASY .I THINK YOU KNOW IT AND DID NOT ASK FOR THE ANSWER.BUT LET US BEGIN WITH THAT .IT WILL SERVE TO OUT LINE THE PROCEDURE AND TO ATTEMPT THE SECOND PROBLEM..
it says that the probability of 2 out of 30 people having the same birthday is 70 percent.
IN THE CONVENTIONAL EASY WAY FOR HIGH SCHOOL STUDENTS UPTO K-12 STANDARD,THIS IS DONE AS FOLLOWS WITH 365 DAYS ONLY CONSIDERED PER YEAR & 30 NUMBER GROUP.
PROBABILITY OF 2 HAVING SAME BIRTH DAY IN THE GROUP = 1 - HAVING NO COINCIDENCE IN BIRTH DAY .....
FOR NO COINCIDENCE TO OCCURR ...............
THE FIRST PERSON CAN TAKE BIRTH ON ANY OF THE 365 DAYS,
THE SECOND PERSON'S BIRTH DAY CAN BE ONLY FROM THE REST 365-1=364 DAYS
THE THIRD PERSON'S BIRTH DAY CAN BE ONLY FROM THE REST 365-2=363 DAYS
....................ETC.....
THE THIRTIETH PERSON'S BIRTH DAY CAN BE ONLY FROM THE REST 365-29=336 DAYS
SO NUMBER OF ALL POSSIBILITIES FOR NO COINCIDENCE =365*364*363*.......*336
TOTAL NUMBER OF POSSIBILITIES WITH NO RESTRICTIONS=365*365*365....30 TIMES =365^30
HENCE PROBABILITY OF NO COINCIDENCE =365*364*363*.......*336/365^30=0.293683757
HENCE PROBABILITY FOR A COINCIDENCE =1- 0.293683757 = 0.706316243..OR ABOUT 70% AS YOU HAVE GOT.
HOWEVER THIS METHOD CAN NOT BE USED FOR 3 OR 4 OR 6 COINCIDENCES AS YOU WANTED FOR SECOND CASE ...
SO LET US EVOLVE A METHOD SUITABLE FOR THAT GENERAL CASE...
I HOPE YOU KNOW PIGEON HOLE PRICIPLE WHICH IS A SIMPLE BUT EFFECTIVE TOOL FOR EXISTENCE PROBLEMS.....
IN SIMPLE TERMS,IT SAYS THAT IF THERE ARE 10 PIGEONS AND 9 CAGES THERE WILL BE ATLEAST ONE CAGE WITH MORE THAN ONE PIGEON WHICH IS OBVIOUSLY CORRECT .
WE WONT USE THE PRINCIPLE DIRECTLY BUT USE ITS ANALOGY TO SOLVE OUR PROBLEM.
SO NOW LET US THINK OF OUR 365 DAYS AS 365 CAGES.30 PERSONS AS 30 PIGEONS.
SO WE WANT OUR CAGES TO BE FILLED AS FOLLOWS FOR OUR FIRST PROBLEM OF ONE COINCIDENCE FOR WHICH WE GOT THE ANSWER OF 70 % NOW.
LET US PUT OUR REQUIREMENT AS NUMBER OF 30 PIGEONS (NUMBER OF PERSONS=30)IN EACH OF 365 CAGES (NUMBER OF DAYS IN AN YEAR = 365 )
here days are different cages and persons are pigeons.
A ...first is CBD-common birthday type cages.
A=[2].TO SHOW THAT WE WANT ONE CAGE WITH 2 PERSONS HAVING SAME BIRTH DAY.
B....second is DBD-different birth day type cages.
B=[1,1,......28 NOS.]THAT IS 28 CAGES WITH 1 PERSON EACH HAVING DIFFERENT BIRTH DAYS OUT OF THE REST 365-1=364 DAYS.
C....third is rest or empty cages.
C=[0,0,....365-30=335 NOS.]THAT IS 335 CAGES WITH NO PERSONS .
THAT IS figures in square brackets [ ] show number of persons in each of different types of cages mentioned above.
[2] indicates 2 persons are there in CBD type.if there are 6 persons with same birth day it will be shown as [6].but,if the 6 are in 3 pairs each with a different common birth day it will be shown as [2,2,2].
LET US SUM IT UP AS DIFFERENT TYPES OF CAGES AS EXPLAINED ABOVE....
A = (1),B =(28),C =(335)..THAT IS 1 CAGE WITH 2 BOYS HAVING SAME BIRTH DAY ,28 CAGES WITH EACH PERSON HAVING DIFFERENT BIRTH DAYS AND 335 CAGES WITH NO PERSONS.
THAT IS figures in parethesis ( ) show the number of types of cages in each catype explained above.(1)under A shows one cage of CBD type.(21)under B indicates 21 cages of DBD type...etc...
we calculate first the ways of selecting the cages (365 days in the year) in to the 3 types ,that is common birth day to be say 1st.jan.,different birthdays to be 2nd.,jan to 30th.jan...…etc...then we find ,the pigeons (30 persons )to be selected to fit those days .we multiply the two to find total number of arrangements possible. we divide it with total of all arrangements without any condition = 365^30 to get the probability.
WITH THIS DISPENSATION WE CAN CALCULATE EACH COMBINATION USING THE FOLLOWING STANDARD FORMULAE..
NUMBER OF COMBINATION OF DIVIDING N THINGS IN TO K GROUPS EACH HAVING SAY X,Y,Z ETC...THINGS SUCH THAT X+Y+Z...=K...IS GIVEN BY N!/X!*Y!*Z!*.....
SO NOW LET US APPLY THIS TO OUR FIRST PROBLEM AND THEN THE SECOND PROBLEM...
*************************************
FIRST PROBLEM.....PROBABILITY OF ATLEAST ONE COINCIDENCE
AS EARLIER THIS IS = 1-PROBABILITY OF NO COINCIDENCE..
B=[1,1....30]
C=[0,0,....335]..OR
HERE A DOES NOT EXIST (0) AS THERE ARE NO COINCIDENCES B=(30),C=(335)..COMBINATION..
1. WE FIRST FIND NUMBER OF SUCH COMBINATIONS POSSIBLE THAT IS 30 OF 1 EACH TYPE AND 335 OF 0 EACH TYPE FROM A TOTAL OF 365....
=365!/(30!*335!)=G SAY
2.NEXT WE FIND THE NUMBER OF POSSIBLE ARRANGEMENTS WITHIN EACH COMBINATION. B= 30!/1!*1!*1!....=30!...
HENCE NO COINCIDENCE COMBINATIONS BY MULTIPLICATION THEOREM IS
#NAME?
PROBABILITY OF NO COINCIDENCE COMBINATIONS =G*B/(365^30)
=365!*30!/(30!*335!*365^30)
=365!/(335!*365^30)=0.293683757
PROBABILITY OF ATLEAST 1 COINCIDENCE =1-G*B/(365^30)=1-0.293683757=0.706316243
OR ABOUT 70 %.=PROBABILITY OF 2 OR MORE HAVING SAME BIRTH DAY.
**********************************************
FIRST PROBLEM WITH A CHANGE .THAT THERE SHOULD BE ONLY 1 COINCIDENCE OR ONLY 2 PEOPLE WILL HAVE SAME BIRTH DAY.USING SAME NOTATION AS ABOVE...
A=[2],B[1,1,....28 CAGES],[0,0,....335 CAGES]
OR...A=(1),B=(28),C=(336)
G=365!/(1!*28!*336!)
A*B=30!/(2!...1 TIME ONLY )*(1!*1!*1!...28 TIMES )=30!/2!
PROBABILITY OF ONLY ONE COINCIDENCE=G*A*B/365^30
=365!*30!/(1!*28!*336!*2!*365^30)
=0.380215027..OR ABOUT 38%
************************************************
NOW THE 6 HAVING SAME BIRTH DAY CAlCULATION
A=[6],B=[1,1,....24 CAGES],C=[0,0,....335 CAGES]
OR...A=(1),B=(24),C=(340)
G=365!/(1!*24!*340!)
A*B=30!/(6!)
PROBABILITY OF 6 COINCIDENCES =(365!*30!)/(1!*24!*340!*6!*365^30)=
Hello,
I have been researching the birthday paradox and it says that the probability of 2 out of 30 people having the same birthday is 70 percent. Then it goes on to say that the probability of 6 out of 30 people having the same birthday is 12 percent. My question is what formulas would I use to find the probablity of 6 out of 30 is 12 percent? This seems too high. Is it even right?
thanks
pierre
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! I have been researching the birthday paradox and it says that the probability of 2 out of 30 people having the same birthday is 70 percent. Then it goes on to say that the probability of 6 out of 30 people having the same birthday is 12 percent. My question is what formulas would I use to find the probablity of 6 out of 30 is 12 percent? This seems too high. Is it even right?
thanks
pierre
OH!THIS IS A RESEARCH PROJECT FOR YOU ..THEN IT IS OK ..
THEN LET US ANALYSE A BIT..THE FIRST PART IS EASY .I THINK YOU KNOW IT AND DID NOT ASK FOR THE ANSWER.BUT LET US BEGIN WITH THAT .IT WILL SERVE TO OUT LINE THE PROCEDURE AND TO ATTEMPT THE SECOND PROBLEM..
it says that the probability of 2 out of 30 people having the same birthday is 70 percent.
IN THE CONVENTIONAL EASY WAY FOR HIGH SCHOOL STUDENTS UPTO K-12 STANDARD,THIS IS DONE AS FOLLOWS WITH 365 DAYS ONLY CONSIDERED PER YEAR & 30 NUMBER GROUP.
PROBABILITY OF 2 HAVING SAME BIRTH DAY IN THE GROUP = 1 - HAVING NO COINCIDENCE IN BIRTH DAY .....
FOR NO COINCIDENCE TO OCCURR ...............
THE FIRST PERSON CAN TAKE BIRTH ON ANY OF THE 365 DAYS,
THE SECOND PERSON'S BIRTH DAY CAN BE ONLY FROM THE REST 365-1=364 DAYS
THE THIRD PERSON'S BIRTH DAY CAN BE ONLY FROM THE REST 365-2=363 DAYS
....................ETC.....
THE THIRTIETH PERSON'S BIRTH DAY CAN BE ONLY FROM THE REST 365-29=336 DAYS
SO NUMBER OF ALL POSSIBILITIES FOR NO COINCIDENCE =365*364*363*.......*336
TOTAL NUMBER OF POSSIBILITIES WITH NO RESTRICTIONS=365*365*365....30 TIMES =365^30
HENCE PROBABILITY OF NO COINCIDENCE =365*364*363*.......*336/365^30=0.293683757
HENCE PROBABILITY FOR A COINCIDENCE =1- 0.293683757 = 0.706316243..OR ABOUT 70% AS YOU HAVE GOT.
HOWEVER THIS METHOD CAN NOT BE USED FOR 3 OR 4 OR 6 COINCIDENCES AS YOU WANTED FOR SECOND CASE ...
SO LET US EVOLVE A METHOD SUITABLE FOR THAT GENERAL CASE...
I HOPE YOU KNOW PIGEON HOLE PRICIPLE WHICH IS A SIMPLE BUT EFFECTIVE TOOL FOR EXISTENCE PROBLEMS.....
IN SIMPLE TERMS,IT SAYS THAT IF THERE ARE 10 PIGEONS AND 9 CAGES THERE WILL BE ATLEAST ONE CAGE WITH MORE THAN ONE PIGEON WHICH IS OBVIOUSLY CORRECT .
WE WONT USE THE PRINCIPLE DIRECTLY BUT USE ITS ANALOGY TO SOLVE OUR PROBLEM.
SO NOW LET US THINK OF OUR 365 DAYS AS 365 CAGES.30 PERSONS AS 30 PIGEONS.
SO WE WANT OUR CAGES TO BE FILLED AS FOLLOWS FOR OUR FIRST PROBLEM OF ONE COINCIDENCE FOR WHICH WE GOT THE ANSWER OF 70 % NOW.
LET US PUT OUR REQUIREMENT AS NUMBER OF 30 PIGEONS (NUMBER OF PERSONS=30)IN EACH OF 365 CAGES (NUMBER OF DAYS IN AN YEAR = 365 )
A=[2].TO SHOW THAT WE WANT ONE CAGE WITH 2 PERSONS HAVING SAME BIRTH DAY.
B=[1,1,......28 NOS.]THAT IS 28 CAGES WITH 1 PERSON EACH HAVING DIFFERENT BIRTH DAYS OUT OF THE REST 365-1=364 DAYS.
C=[0,0,....365-30=335 NOS.]THAT IS 335 CAGES WITH NO PERSON LEFT .
LET US SUM IT UP AS DIFFERENT VARIETY OF CAGES THAT IS BASED ON NUMBER OF PERSONS IN THE CAGE...A = (1),B =(28),C =(335)..THAT IS 1 CAGE WITH 2 BOYS HAVING SAME BIRTH DAY ,28 CAGES WITH 1 PERSON HAVING DIFFERENT BIRTH DAYS AND 335 CAGES WITH NO PERSONS.
WITH THIS DISPENSATION WE CAN CALCULATE EACH COMBINATION USING THE FOLLOWING STANDARD FORMULAE..
NUMBER OF COMBINATION OF DIVIDING N THINGS IN TO K GROUPS EACH HAVING SAY X,Y,Z ETC...THINGS SUCH THAT X+Y+Z...=K...IS GIVEN BY N!/X!*Y!*Z!*.....
SO NOW LET US APPLY THIS TO OUR FIRST PROBLEM AND THEN THE SECOND PROBLEM...
*************************************
FIRST PROBLEM.....PROBABILITY OF ATLEAST ONE COINCIDENCE
AS EARLIER THIS IS = 1-PROBABILITY OF NO COINCIDENCE..
B=[1,1....30]
C=[0,0,....335]..OR
HERE A DOES NOT EXIST (0) AS THERE ARE NO COINCIDENCES B=(30),C=(335)..COMBINATION..
1. WE FIRST FIND NUMBER OF SUCH COMBINATIONS POSSIBLE THAT IS 30 OF 1 EACH VARIETY AND 335 OF 0 EACH VARIETY FROM A TOTAL OF 365....
=365!/(30!*335!)=G SAY
2.NEXT WE FIND THE NUMBER OF POSSIBLE ARRANGEMENTS WITHIN EACH COMBINATION. B= 30!/1!*1!*1!....=30!...
HENCE NO COINCIDENCE COMBINATIONS BY MULTIPLICATION THEOREM IS
=G*B
PROBABILITY OF NO COINCIDENCE COMBINATIONS =G*B/(365^30)
=365!*30!/(30!*335!*365^30)
=365!/(335!*365^30)=0.293683757
PROBABILITY OF ATLEAST 1 COINCIDENCE =1-G*B/(365^30)=1-0.293683757=0.706316243
OR ABOUT 70 %.=PROBABILITY OF 2 OR MORE HAVING SAME BIRTH DAY.
**********************************************
FIRST PROBLEM WITH A CHANGE .THAT THERE SHOULD BE ONLY 1 COINCIDENCE OR ONLY 2 PEOPLE WILL HAVE SAME BIRTH DAY.USING SAME NOTATION AS ABOVE...
A=[2],B[1,1,....28 CAGES],[0,0,....335 CAGES]
OR...A=(1),B=(28),C=(336)
G=365!/(1!*28!*336!)
A*B=30!/(2!...1 TIME ONLY )*(1!*1!*1!...28 TIMES )=30!/2!
PROBABILITY OF ONLY ONE COINCIDENCE=G*A*B/365^30
=365!*30!/(1!*28!*336!*2!*365^30)
=0.380215027..OR ABOUT 38%
************************************************
NOW I LEAVE THE 6 HAVING SAME BIRTH DAY CAlCULATION TO YOU WITH BASIS REPEATED.
A=[6],B=[1,1,....24 CAGES],C=[0,0,....335 CAGES]
OR...A=(1),B=(24),C=(340)
|
|
|
