The toughest question for any statistician yet!?

a) If there are numbers between 1 and 49, and 6 are to be drawn at random, what are the possible number of outcomes?


b) Imagine if I write 1,2,3,4,5 and 6 on a note book, and 1 - 49 written on small pieces (i.e. 49 small pieces of paper of paper) kept in a box, what is the probability that the 6 numbers i draw from the box are:


i) the same 6 number


ii) 5 matching numbers


iii) 4 matching numbers


iv) 3 matching numbers


v) 2 matching numbers


vi) 1 matching numbers


vii) no matching numbers





C) does the number of possible outcomes have an effect on the probability of matching numbers?





P.S. numbers once drawn cannot be repeated and I know it is not a permutation question. so 49P6 is not the right answer for the first question.





please give formulae aswell

The toughest question for any statistician yet!?
a) It is simply 49C6 = 13,983,816





b i) The prob that those 6 numbers are 1-6 is simply


1/13,983,816


since only one (6C6) of those combinations involve 1-6.


Alternatively you could do:


1/49 * 1/48 * 1/47 * 1/46 * 1/45 * 1/44 * 6!


= 1/13,983,816





ii) 5 matching numbers.


There are 6C5 ways of choosing 5 of the 6 nos, and 43C1 = 43 ways of choosing the other one.


That is 258 possibilities


P = 258 / 13,983,816





iii) 4 matching


No. of ways = 6C4 * 43C2 = 13545


P = 13545 / 13,983,816





Generally n matching nos:


No of ways = 6Cn * [43C(6-n)]


Total cases = 49C6


P(n matching) = 6Cn * [43C(6-n)] / 49C6





You can use this formula to work out all of part b, and verify that they all add up to 1.
Reply:PEANUT BUTTER JELLY


really hard dude...bt vil luv to know the answer...
Reply:Canada has a 6/49 Lottery which works somewhat as what you try to describe. The numbers are picked at random, without replacement.





If the order of the numbers is important (it is not in the lottery, but let us start there):


The first number drawn can be any one of 49.


The second number can be any of remaining 48.


...


The sixth number can be any one of remaining 44.





The number of possible outcomes is:


49*48*47*46*45*44 = 10,068,347,520





However, if the order in which the numbers come out does not matter, then there are 6*5*4*3*2*1 ways that the same 6 numbers can come out (the first one can be any of the six, the second can be any of the remaining five...). That is 720 ways.





10,068,347,520 / 720 = 13,983,816


which is C(49,6) or 49P6 (if that means "49 pick 6").


It is 49! / [(49-6)! * 6!]


where ! is "factorial": the product of all numbers from 1 to the number.





So, if you seek to have the same 6 numbers, then the probability is 1 in 13,983,816





This particular calculation has been done often (because of the 6/49 lottery). it can be found on the web.
Reply:You are going to play lotto,? To get all 6 it's about


13000000, to 1, off the top of my head. it's better than Jelly.EDIT-he is talking about matching #s to the ones picked ,Like lotto.


I think you multiply 46x6, six times.46x6=276, 276x6=


not positive.
Reply:Well, I'm no statician, I can't even spell it. But since no numbers can be repeated then i-v have a zero possibility. vi--I have no idea what one matching number means. Vii would be a 100% chance because there are no matching numbers. I don't know of the possible number of outcomes, but I don't see how you will have ANY matching numbers given the criteria you've provided.