Question 32723: Suppose 25 people are arranged in a circle. Three of them are selected at a random. what is the probability that none of those selected were next to each other?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Suppose 25 people are arranged in a circle. Three of them are selected at a random. what is the probability that none of those selected were next to each other?
KEY TO SOLVING THESE PROBLEMS IS TO DEVELOP A LOGICAL PROCEDURE WHICH
1.IS AMENABLE TO CALCULATION WITH KNOWN/STANDARD FORMULAE OR FROM FUNDAMENTALS
2.SATISFIES ALL POSITIVE REQUIREMENTS OF THE PROBLEM
3.EXCLUDES ALL NEGATIVE REQUIREMENTS OR EXCEPTIONS IN THE PROBLEM.
4.NO DUPLICTION OR OVERLAP OF THE DONE THINGS OCCUR AT ANY STAGE IN THE SOLUTION
THESE PROBLEMS ARE MANY TIMES TRICKY OR DECEPTIVE IN THAT THEY GIVE US A FALSE SENSE OF CORRECTNESS AND SOME FLAW IN OUR LOGIC IS NOT SHOWN UP TILL A LATER DATE WHEN WE REALSE IT OUR SELVES OR SOME ONE ELSE POINTS OUT.
HENCE THE BEST WAY TO LEARN THESE THINGS IS TO HAVE AN OPEN MIND IN OUR APPROACH,DISCUSS THE LOGIC WITH OTHERS ACCEPTING THEIR SUGGESTIONS`ON ANY OF OUR FLAWS AND TO CONTINUOUSLY KEEP IMPROVING BY CORRECTING OUR MISTAKES.
IN ONE WORD SAY "WE WELCOME CRITICISM"..THE SAME THING APPLIES TO ME TOO!FEEL FREE TO POINT OUT IF YOU FIND ANY LOGIC NOT CLEAR OR DOUBT FULL.WE OWE YOU A THOUSAND THANKS FOR EACH FLAW POINTED OUT IN OUR LOGIC.
SO NOW ON TO THE PROBLEM....
1.LET US NAME THE PEOPLE FOR CONVENIENCE AS P1,P2.......P25 IN THEIR ORDER OF ORIGINAL ARRANGEMENT IN A CIRCLE.
2.TOTAL NUMBER OF COMBINATIONS OF 3 PEOPLE FROM THESE 25 ARE 25C3=25!/(3!*22!)
3.WE HAVE TO FIND THE NUMBER OF SELECTIONS WHERE NONE OF THOSE SELECTED WERE NEXT TO EACH OTHER.
4.NOW LET US SAY WE PICKED ONE PERSON FIRST..THIS CAN BE DONE IN 25 WAYS.
5.LET US SAY THE PERSON SELECTED IS P1
6.THEN THE NEXT PERSON CAN BE FROM PEOPLE OTHER THAN P2 AND P24,AS THEY ARE NEIGHBORS TO P1.SO WE HAVE 25-P1-P2-P25=22 PEOPLE TO CHOSE FROM FOR THE SECOND PERSON IN 22 WAYS.
7.LET US SPLIT THESE 22 POSSIBILITIES INTO 3 CASES AS FOLLOWS OR REASON YOU WOULD KNOW SOON.
8.CASE 1....P3 IS SELECTED....IN 1 WAY..
9.CASE 2....P24 IS SELECTED...IN 1 WAY..
10.CASE 3....ANY OTHER PERSON IS SELECTED...IN 20 WAYS.
11.NOW IN CASE 1. THE THIRD PERSON BY THE SAME LOGIC (NO.6) AS ABOVE CAN BE FROM PEOPLE OTHER THAN P4,P2,P1,P25,P3....THAT IS 25-5=20 WAYS.
12.SIMILARLY IN CASE 2. THE THIRD PERSON BY THE SAME LOGIC (NO.6) AS ABOVE CAN BE FROM PEOPLE OTHER THAN P23,P25,P1,P2,P24....THAT IS 25-5=20 WAYS.
13.BUT IN THE THIRD CASE THEY CAN BE ANY ONE FROM 25-P1-P2-P25-PSECOND PICK -2 PERSONS ON EITHER SIDE OF SECOND PICK=19 WAYS.
14. (I THINK NOW IT IS CLEAR WHY WE MADE THE SPLIT INTO 3 CASES)
15.SO TOTAL NUMBER OF ARRANGEMENTS ARE
25*{1*20+1*20+20*19)=25*21*20....BUT THESE ARE ARRANGEMENTS...SO NUMBER OF COMBINATIONS ARE 25*21*20/3!
PROBABILITY THAT NONE OF THOSE SELECTED WERE NEXT TO EACH OTHER ORIGINALL=
={25*21*20/3!}/(25!/(3!*22!)
=25*21*20*3!/25*24*23*3!
=(21*20)/(24*23)=
=0.76
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